Answer:
The product represent the number [tex]3\times 0.3=0.9[/tex]
Step-by-step explanation:
To find : The decimal 0.3 represents What type of number best describes 0.9, which is 3.0.3?
Solution :
0.3 represents [tex]0.3=\frac{3}{10}[/tex]
0.9 represents [tex]0.9=\frac{9}{10}[/tex]
If we multiply 0.3 by 3 we get 0.9
As, [tex]3\times 0.3=3\times \frac{3}{10}[/tex]
[tex]3\times 0.3=\frac{9}{10}[/tex]
[tex]3\times 0.3=0.9[/tex]
Therefore, The product represent the number [tex]3\times 0.3=0.9[/tex]
How many subsets does the set D={c,a,t}D={c,a,t} have?
Answer:
The number of all possible subsets of D is 8.
Step-by-step explanation:
Consider the provided set D={c,a,t}
The subset of D contains no elements: { }
The subset of D contains one element each: {c} {a} {t}
The subset of D contains two elements each: {c, a} {a, t} {c, t}
The subset of D contains three elements: {c, a, t)
Hence, all possible subsets of D are { }, {c}, {a}, {t}, {c, a}, {a, t}, {c, t}, {c, a, t}
Therefore, number of all possible subsets of D is 8.
Or we can use the formula:
The number of subsets of the set is [tex]2^n[/tex] If the set contains ‘n’ elements.
There are 3 elements in the provided set, thus use the above formula as shown:
2³=8
Hence, the number of all possible subsets of D is 8.
Prove: If n is a positive integer andn2 is
divisible by 3, then n is divisible by3.
Answer and Step-by-step explanation:
n > 0
n² divisible by 3 ⇒ n is divisible by 3.
Any number divisible by 3 has the sum of their components divisible by 3.
If n² is divisible by 3, we can say that n² can be written as 3*x.
n² = 3x ⇒ n = √3x
As n is a positive integer √3x must be a integer and x has to have a 3 factor. (x = 3.a.b.c...)
This way, we can say that x = 3y and y is a exact root, because n is a integer.
n² = 3x ⇒ n = √3x ⇒ n = √3.3y ⇒ n = √3.3y ⇒ n = √3²y ⇒ n = 3√y
Which means that n is divisible by 3.
The population of Cook Island was always been modeled by a logistic equation with growth rate r=19 and carrying capacity N=8000, with time t measured in years. However, beginning in 2000, 9 citizens of Cook Island have left every year to become a mathematician, never to return. Find the new differential equation modeling the population of the island P(t) after 2000. Use P for P(t) and P' for P′(t)
The answer is P' = P/9(1-P/8000)-9
Answer:
[tex]P'(t) = 19P(1 - \frac{P}{8000}) - 9[/tex]
Step-by-step explanation:
The logistic equation is given by Equation 1):
1) [tex]\frac{dP}{dt} = rP(1 - \frac{P}{N})[/tex]
In which P represents the population, [tex]\frac{dP}{dt} = P'(t)[/tex] is the variation of the population in function of time, r is the growth rate of the population and N is the carrying capacity of the population.
Now for your system:
The problem states that the population has growth rate r=19.
The problem also states that the population has carrying capacity N=8000.
We can replace these values in Equation 1), so:
[tex]P'(t) = 19P(1 - \frac{P}{8000})[/tex]
However, beginning in 2000, 9 citizens of Cook Island have left every year to become a mathematician, never to return. So, we have to subtract these 9 citizens in the P'(t) equation. So:
[tex]P'(t) = 19P(1 - \frac{P}{8000}) - 9[/tex]
The correct differential equation modeling the population of Cook Island after 2000, taking into account the emigration of 9 citizens every year, is given by:[tex]\[ P' = \frac{P}{9}\left(1 - \frac{P}{8000}\right) - 9 \][/tex]
To derive this equation, we start with the standard logistic growth model, which is given by:
[tex]\[ P' = rP\left(1 - \frac{P}{K}\right) \][/tex]
where \( r \) is the intrinsic growth rate and [tex]\( K \)[/tex] is the carrying capacity of the environment. For the Cook Islands, we have [tex]\( r = 19 \) and \( K = 8000 \)[/tex].
However, since 9 citizens leave the island every year starting from 2000, we need to modify the logistic growth model to account for this emigration. The term representing the natural growth of the population remains the same, but we subtract 9 from the growth rate to represent the annual emigration:
[tex]\[ P' = rP\left(1 - \frac{P}{K}\right) - 9 \][/tex]
Substituting the given values of [tex]\( r \)[/tex] and [tex]\( K \)[/tex] into the equation, we get:
[tex]\[ P' = 19P\left(1 - \frac{P}{8000}\right) - 9 \][/tex]
Now, we need to adjust the growth rate [tex]\( r \)[/tex] to reflect the fact that the population is also decreasing due to emigration. Since the population decreases by 9 every year, we divide the growth rate by 9 to account for this decrease:
[tex]\[ P' = \frac{19P}{9}\left(1 - \frac{P}{8000}\right) - 9 \][/tex]
However, the growth rate should not be divided by 9, as this would incorrectly alter the per capita growth rate. The correct adjustment is to subtract the constant rate of emigration from the overall growth rate:
[tex]\[ P' = 19P\left(1 - \frac{P}{8000}\right) - 9 \][/tex]
Upon reviewing the provided answer, we see that the growth rate [tex]\( r \)[/tex]has been incorrectly divided by 9. The correct differential equation should not have the growth rate divided by 9. Therefore, the correct differential equation modeling the population of the island [tex]\( P(t) \)[/tex] after 2000 is:
[tex]\[ P' = 19P\left(1 - \frac{P}{8000}\right) - 9 \][/tex]
Find the average cost per item when the required number of items are produced. C(x) = 19x + 1900, 1000 items What is the average cost per item?
Answer:
$20.9
Step-by-step explanation:
We have been given a formula [tex]C(x)=19x+1900[/tex], which represents the cost of x items.
First of all, we will find cost of 1000 items by substituting [tex]x=1000[/tex] in our given formula as:
[tex]C(1000)=19(1000)+1900[/tex]
[tex]C(1000)=19,000+1900[/tex]
[tex]C(1000)=20,900[/tex]
To find average cost per item, we will divide total cost by number of items as:
[tex]\text{Average cost per item}=\frac{\$20,900}{1000}[/tex]
[tex]\text{Average cost per item}=\$20.9[/tex]
Therefore, the average cost per item would be $20.9.
What is the principal square root of -4
Answer:
The principal square root of -4 is 2i.
Step-by-step explanation:
[tex]\sqrt{-4}[/tex] = 2i
We have the following steps to get the answer:
Applying radical rule [tex]\sqrt{-a} =\sqrt{-1} \sqrt{a}[/tex]
We get [tex]\sqrt{-4} =\sqrt{-1} \sqrt{4}[/tex]
As per imaginary rule we know that [tex]\sqrt{-1}=i[/tex]
= [tex]\sqrt{4} i[/tex]
Now [tex]\sqrt{4} =2[/tex]
Hence, the answer is 2i.
You go to a car dealer and pick out a vehicle that costs $31,210 "out-the-door." Instead of paying all the cash upfront, you can put down an amount and finance the rest of the car loan. The money will be financed over 5 years at 4.5%. By hand, compute the following:
If you put down 15% of the car’s cost, what is the amount of the car loan?
Find the total amount paid for the car (including the down payment)
Answer:
Cost of car = $31,210
Now we are given that you put down 15% of the car’s cost.
So, Down payment = [tex]15\% \times 31210[/tex]
= [tex]\frac{15}{100} \times 31210[/tex]
= [tex]4681.5[/tex]
So, Amount of car loan = Total cost - Down payment
Amount of car loan =$31210 - $4681.5
=$26528.5
Thus Amount of car loan is $26528.5
Now To find the total amount of car
Principal = $26528.5
Rate of interest = 4.5%
Time = 5 years
[tex]A=P(1+r)^t[/tex]
[tex]A=26528.5(1+\frac{4.5}{100})^5[/tex]
[tex]A=33059.337533[/tex]
Total amount including down payment = $33059.337533+$4681.50 = $37740.837533
Hence the total amount paid for the car (including the down payment) is $37740.83
Let 'A' and 'B' be subsets of a universal set 'U'. 1. Which of the following describes 'A cap B"?! • 1. The set of all elements of 'Uʻ that are elements of both 'A' and 'B'. • 2. The set of all elements of 'U' that are elements of either 'A' or 'B 3. The set of all elements of ‘Uʻ that are elements of neither ‘A’ nor 'B'. • 4. The set of all elements of 'U' that are elements of either 'A' or 'B' but not both. 2. Which of the following describes '(A cup B)""? • 1. The set of all elements of 'U' that are elements of both 'A' and 'B'. 2. The set of all elements of 'U' that are elements of either 'A' or 'B'. • 3. The set of all elements of 'U' that are elements of neither 'A' nor 'B'. • 4. The set of all elements of 'U' that are elements of either 'A' or 'B' but not both.
Answer:
A ∩ B: 1. The set of all elements of 'Uʻ that are elements of both 'A' and 'B'.
A ∪ B: 2. The set of all elements of 'U' that are elements of either 'A' or 'B'.
Step-by-step explanation:
1. The "intersection" symbol (∩) signifies the members that are in both sets. For example, {1, 2} ∩ {1, 3} = {1}.
__
2. The "union" symbol (∪) signifies the members that are in either set. For example, {1, 2} ∪ {1, 3} = {1, 2, 3}.
Use properties of limits and algebraic methods to find the limit, if it exists. (If the limit is infinite, enter '[infinity]' or '-[infinity]', as appropriate. If the limit does not otherwise exist, enter DNE.) lim x→3 f(x),
where f(x) = 9 − 3x if x < 3 ;
and x^2 − x if x ≥ 3
Answer:
The limit of this function does not exist.
Step-by-step explanation:
[tex]\lim_{x \to 3} f(x)[/tex]
[tex]f(x)=\left \{ {{9-3x} \quad if \>{x \>< \>3} \atop {x^{2}-x }\quad if \>{x\ \geq \>3 }} \right.[/tex]
To find the limit of this function you always need to evaluate the one-sided limits. In mathematical language the limit exists if
[tex]\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) =L[/tex]
and the limit does not exist if
[tex]\lim_{x \to a^{-}} f(x) \neq \lim_{x \to a^{+}} f(x)[/tex]
Evaluate the one-sided limits.
The left-hand limit
[tex]\lim_{x \to 3^{-} } 9-3x= \lim_{x \to 3^{-} } 9-3*3=0[/tex]
The right-hand limit
[tex]\lim_{x \to 3^{+} } x^{2} -x= \lim_{x \to 3^{+} } 3^{2}-3 =6[/tex]
Because the limits are not the same the limit does not exist.
If you roll one die and flip one coin, what are all the possible outcomes?
Answer: [tex](1,T), (2,T), (3,T), (4, T), (5,T), (6,T)\\(1,H), (2,H), (3,H), (4, H), (5,H), (6,H)[/tex]
Step-by-step explanation:
The total outcomes on a die = {1,2,3,4,5,6}=6
The total outcomes on a coin = {Tails or Heads}=2
The number of possible outcomes =[tex]6\times2=12[/tex]
If you roll one die and flip one coin, then the possible outcomes are:
[tex](1,T), (2,T), (3,T), (4, T), (5,T), (6,T)\\(1,H), (2,H), (3,H), (4, H), (5,H), (6,H)[/tex]
Here T denotes for Tails and H denotes for heads.
9 + 22 = x + 1
HALPP
Answer:
x = 30
Step-by-step explanation:
9 + 22 = x + 1
9 + 22 = 31
31 = x + 1
-1 -1
30 = x
x = 30
The dimensions of a nicotine transdermal patch system are 4.7 cm by 4.8 cm. Express these dimensions in corresponding inches if 1 inch is equivalent to 25.4 mm.
Answer:
1.85 inches by 1.89 inches.
Step-by-step explanation:
We have been given that the dimensions of a nicotine transdermal patch system are 4.7 cm by 4.8 cm.
First of all, we will convert given dimensions into mm.
1 cm equals 10 mm.
4.7 cm equals 47 mm.
4.8 cm equals 48 mm.
We are told that 1 inch is equivalent to 25.4 mm, so to find new dimensions, we will divide each dimension by 25.4 as:
[tex]\frac{47\text{ mm}}{\frac{25.4\text{ mm}}{\text{inch}}}=\frac{47\text{ mm}}{25.4}\times \frac{\text{ inch}}{\text{mm}}=1.85039\text{ inch}\approx 1.85\text{ inch}[/tex]
[tex]\frac{48\text{ mm}}{\frac{25.4\text{ mm}}{\text{inch}}}=\frac{48\text{ mm}}{25.4}\times \frac{\text{ inch}}{\text{mm}}=1.8897\text{ inch}\approx 1.89\text{ inch}[/tex]
Therefore, the corresponding dimensions would be 1.85 inches by 1.89 inches.
Final answer:
To convert the dimensions of a nicotine transdermal patch from centimeters to inches, multiply the centimeter measurements by 10 to get millimeters, and then divide by 25.4 to get inches. The patch measures approximately 1.85 inches by 1.89 inches.
Explanation:
The student is asking to convert the dimensions of a nicotine transdermal patch system from centimeters to inches.
Given that 1 inch equals 25.4 millimeters (mm), this can be done by first converting the dimensions from centimeters (cm) to millimeters and then from millimeters to inches.
Since 1 cm equals 10 mm, the dimensions of the patch in millimeters are 47 mm by 48 mm. To convert these dimensions to inches, we would divide each by 25.4 (since there are 25.4 mm in an inch).
So, the dimension in inches for the patch's length would be 47 mm / 25.4 mm/inch ≈ 1.85 inches, and its width would be 48 mm / 25.4 mm/inch ≈ 1.89 inches.
Therefore, the nicotine patch measures approximately 1.85 inches by 1.89 inches.
If the area of polygon P is 72 square units, what is the scaled factor did deigo use to go from P to Q?
Answer:
The scale factor used to go from P to Q is 1/4
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
If two figures are similar, then the ratio of its areas i equal to the scale factor squared
Let
z ----> the scale factor
x -----> area of polygon Q
y -----> area of polygon P
[tex]z^{2}=\frac{x}{y}[/tex]
we have
[tex]y=72\ units^2[/tex]
Find the area of polygon Q
Divide the the area of polygon Q in two triangles and three squares
The area of the polygon Q is equal to the area of two triangles plus the area of three squares
see the attached figure N 2
Find the area of triangle 1
[tex]A=(1/2)(1)(2)=1\ units^2[/tex]
Find the area of three squares (A2,A3 and A4)
[tex]A=3(1)^2=3\ units^2[/tex]
Find the area of triangle 5
[tex]A=(1/2)(1)(1)=0.5\ units^2[/tex]
The area of polygon Q is
[tex]x=1+3+0.5=4.5\ units^2[/tex]
Find the scale factor
[tex]z^{2}=\frac{x}{y}[/tex]
we have
[tex]y=72\ units^2[/tex]
[tex]x=4.5\ units^2[/tex]
substitute and solve for z
[tex]z^{2}=\frac{4.5}{72}[/tex]
[tex]z^{2}=\frac{1}{16}[/tex]
square root both sides
[tex]z=\frac{1}{4}[/tex]
therefore
The scale factor used to go from P to Q is 1/4
The scaled factor did Diego used to go from P to Q is 1/4.
Given
The area of polygon P is 72 square units.
What is the scale factor?If two figures are similar, then the ratio of its areas is equal to the scale factor squared.
[tex]\rm z^2=\dfrac{x}{y}[/tex]
Where; z is the scale factor, x = area of polygon Q, y = area of polygon P.
Therefore,
The area of the first triangle is;
[tex]\rm Area = \dfrac{1}{2} \times base \times height\\\\Area = \dfrac{1}{2} \times 1 \times 2\\\\Area = 1 \ units[/tex]
The area of three squares (A2, A3, and A4)
[tex]\rm Area = 3(1)^2\\\\Area = 3\ square \ units[/tex]
The area of the 5th triangle is;
[tex]\rm Area = \dfrac{1}{2} \times base \times height\\\\Area = \dfrac{1}{2} \times 1 \times 1\\\\Area = 0.5 \ units[/tex]
Then,
The area of the polygon is;
x = 1 + 3 + 0.5 = 4.5 units
Therefore,
The scaled factor did Diego used to go from P to Q is;
[tex]\rm z^2=\dfrac{4.5}{72}\\\\z^2=\dfrac{1}{16}\\\\z=\dfrac{1}{4}\\\\[/tex]
Hence, the scaled factor did Diego used to go from P to Q is 1/4.
To know more about the Scale factor click the link given below.
https://brainly.com/question/22644306
Determine the payment to amortize the debt. (Round your answer to the nearest cent.) Quarterly payments on $19,500 at 3.9% for 6 years.
Answer:
$925.20
Step-by-step explanation:
Loan Amount, P = $19,500
Rate of interest, r = 3.9%
Time, t = 6 years
Payment mode, n = Quarterly (4)
payment to amortize, EMI = ?
Formula: [tex]EMI=\dfrac{P\cdot \frac{r}{n}}{1-(1+\frac{r}{n})^{-n\cdot t}}[/tex]
where,
n = 4 , Rate of interest , r = 0.039
Put the values into formula
[tex]EMI=\dfrac{19500\cdot \frac{0.039}{4}}{1-(1+\frac{0.039}{4})^{-4\cdot 6}}[/tex]
[tex]EMI=915.20[/tex]
Hence, The payment to amortize the debt is $915.20
A and B are bounded non-empty subsets of R. For inf(A) to be less than or equal to inf(B), which of the following conditions must be met?
a) For every b in B and epsilon > 0, there exists a in A, such that a < b + epsilon.
b) There exists a in A, and b in B such that a < b.
If neither of these conditions are appropriate, what would be appropriate conditions for inf(A) to be less than or equal to inf(B)?
Answer:
a) must be met
Step-by-step explanation:
We have two conditions:
a) For every [tex]b\in B[/tex] and [tex]\epsilon>0[/tex], there exists [tex]a\in A[/tex], such that [tex]a<b+\epsilon[/tex].
b) There exists [tex]a\in A[/tex] and [tex]b\in B[/tex] such that [tex]a<b[/tex].
We will prove that conditon a) is equivalent to [tex]inf(A)\leq inf(B)[/tex]
If a) is not satisfied, then it would exist [tex]b\in B[/tex] and [tex]\epsilon >0[/tex] such that, for every [tex]a\in A[/tex], [tex]a\geq b+\epsilon[/tex]. This implies that [tex]b+\epsilon[/tex] is a lower bound for A and in consequence
[tex]inf(A)\geq b+\epsilon > b\geq inf(B)[/tex]
Then, [tex]inf(A) \leq inf(B)[/tex] implies a).
If [tex]inf(A) \leq inf(B)[/tex] is not satisfied then, [tex]inf(A) > inf(B)[/tex] and in consequence exists [tex]b\inB[/tex] such that [tex]b-inf(A)=\epsilon >0[/tex]. Then [tex]b-\epsilon=inf(A)[/tex] and, for every [tex]a\in A[/tex],
[tex]b-\epsilon =inf(A)\leq a[/tex].
So, a) is not satisfied.
In conclusion, a) is equivalent to [tex]inf(A)\leq inf(B)[/tex]
Finally, observe that condition b) is not an appropiate condition to determine if [tex]inf(A)\leq inf(B)[/tex] or not. For example:
A={0}, B={1}. b) is satisfied and [tex]inf(A)=0<1=inf(B)[/tex]A={0}. B={-1,1}. b) is satisfied and [tex]inf(A)=0>-1=inf(B)[/tex]The "absorption law" (theorem 2.1.1 in our book) states that p V (p Aq) is logically equivalent to p. Construct a truth table to show these statements are equivalent.
Answer:
According to the Law of Absorption, these 2 expressions are equivalent:
p ∨ (p ∧ q) = p
Truth Table:
(see the image attached)
Step-by-step explanation:
To construct the Truth Table you can consider the 4 possible combinations of states that p and q could have, that is
1. p=T, q=T
2. p=T, q=F
3. p=F, q=T
4. p=F, q=F
Then you can calculate p ∨ (p ∧ q) = p for each combination
1. T ∨ (T ∧ T) = T
2. T ∨ (T ∧ F) = T
3. F ∨ (F ∧ T) = F
4. F ∨ (F ∧ F) = F
You can see that the previous values are the same states that p has, you can also see it in the table attached.
Please help me with this question.
Will mark brainliest
Thanks so much
Answer:
☑ 30y²
☑ 30y² + x
Step-by-step explanation:
Polynomials contain indeterminates [variables] and operation performances, non-including negative exponents, fractional exponents, etcetera.
I am joyous to assist you anytime.
The marketing department at Cable TV (CTV) wants to know how promotional advertising affects the number of viewers for the Saturday Night Movie. Research shows that 10 million viewers watched the movie when CTV ran 15 one-minute ads on Friday. When they ran 25 one-minute ads on Friday, the movie had 18 million viewers. Use linear interpolation to estimate the number of viewers if CTV runs 23 one-minute ads on Friday.
Answer:
16.4 million viewers
Step-by-step explanation:
The number of viewers increased by 8 million from 10 to 18 million when the number of ads increased by 10 ads from 15 to 25. If 23 ads are run, that represents an increase of 8 ads from 15, so we expect 8/10 of the increase in viewers.
8/10 × 8 million = 6.4 million
The number we expect with 23 ads is 6.4 million more viewers than 10 million viewers, so is 16.4 million.
_____
Alternate solution
We can write a linear equation in 2-point form for the number of viewers expected for a given number of ads:
y = (18 -10)/(25 -15)(x -15) +10
y = (8/10)(x -15) +10
y = 0.8x -2 . . . . . million viewers for x ads
For 23 ads, this gives ...
y = 0.8×23 -2 = 18.4 -2 = 16.4 . . . . million viewers, as above
_____
Comment on 8/10
I consider it coincidence that the number 23 is 8/10 of the difference between 25 and 15, and the slope of the line is 8/10. The point we're trying to interpolate has no relationship to the slope of the line, and vice versa.
Linear interpolation illustrates the use of linear equation of several points
The number of viewers is 16.4 million, if a 23 one-minute ads runs on Friday.
Linear interpolation is represented as:
[tex]\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}[/tex]
Let:
[tex]x \to[/tex] Time
[tex]y \to[/tex] Viewers
So, we have:
[tex](x_1,y_1) = (15,10m)[/tex]
[tex](x_2,y_2) = (25,18m)[/tex]
[tex](x,y) = (23,y)[/tex]
Substitute the above points in:
[tex]\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}[/tex]
So, we have:
[tex]\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}[/tex]
[tex]\frac{18m - 10m}{25 -15} = \frac{y - 10m}{23 -15}[/tex]
[tex]\frac{8m}{10} = \frac{y - 10m}{8}[/tex]
Multiply both sides by 8
[tex]\frac{64m}{10} = y - 10m[/tex]
[tex]6.4m = y - 10m[/tex]
Collect like terms
[tex]y =10m + 6.4m[/tex]
[tex]y =16.4m[/tex]
Hence, the number of viewers is 16.4 million, if a 23 one-minute ads runs on Friday.
Read more about linear interpolation at:
https://brainly.com/question/4248868
What is the area under the curve y=x−x^2and above the x-axis?
Answer:
The area between the x-axis and the given curve equals 1/6 units.
Step-by-step explanation:
given any 2 functions f(x) and g(x) the area between the 2 figures is calculated as
[tex]A=\int_{x_1}^{x_2}(f(x)-g(x))dx[/tex]
The area needed is shown in the attached figure
The points of intersection of the given curve and x-axis are calculated as
[tex]x-x^2=0\\\\x(1-x)=0\\\\\therefore x=0,x=1[/tex]
hence the points of intersection are[tex](0,0),(1,0)[/tex]
The area thus equals
[tex]A=\int_{0}^{1}(x-x^2-0)dx\\\\A=\int_{0}^{1}xdx-\int_{0}^{1}x^2dx\\\\A=1/2-1/3\\\\A=1/6[/tex]
find the unpaid balance on the debt after 5 years of monthly payments on $190,000 at 3% for 25 years
Answer:
the unpaid balance after the 5 years will be 125400.
Step-by-step explanation:
Given,
Principal amount, P = 190,000
rate,r = 3%
total time,t = 25 years
So, the total interest after 25 years will be,
[tex]I\ =\ \dfrac{P\times r\times t}{100}[/tex]
[tex]=\ \dfrac{190,000\times 3\times 25}{100}[/tex]
= 142500
amount will be paid in 3 years with same interest rate can be given by
[tex]I_p\ =\ \dfrac{P\times r\times t}{100}[/tex]
[tex]=\ \dfrac{190,000\times 3\times 3}{100}[/tex]
= 17100
So, the amount of interest to be paid= 142500 - 17100
= 125400
so, the unpaid amount of interest after the 5 years will be 125400.
solve each equation with steps.
6r+7=13+7
develop an explicit formula in terms of n for the nth term of the following sequence:
0,4,18,48,100,180,......
show that it "works" for the sixth term and use it to find the seventh term.
The given sequence : 0,4,18,48,100,180,......
We can write the terms of the sequence as :
[tex]\text{Ist term }:a_1=1(1^2-1)=0\\\\\text{IInd term }: a_2=2^2(2-1)=4(2-1)=4\\\\\text{IIIrd term }:a_3=3^2(3-1)=9(2)=18\\\\\text{IVth term }:a_4=4^2(4-1)=(16)(3)=48\\\\\text{Vth term }:a_5=5^2(5-1)=25(4)=100\\\\\text{VIth term }:a_6=6^2(6-1)=36(5)=180[/tex]
From the above presentation of the terms, the explicit formula in terms of n for the nth term will be :-
[tex]a_n=n^2(n-1)[/tex]
Put n= 7 , we get
[tex]a_7=7^2(7-1)=49(6)=294[/tex]
Therefore, the seventh term of the given sequence = 294
On three examinations, you have grades of 85, 78, and 84. There is still a final examination, which counts as one grade In order to get an A your average must be at least 90. If you get 100 on the final, what is your numerical average? 86.75 (Type an integer or a decimal.) Is an A in the course possible? Yes No To eam a B in the course, you must have a final average of at least 80 What grade must you got on the final to earn a B in the course? Type an integer or a decimal) Enter your answer in the answer box and then click Check Answer All parts showing Clear All Pearson Copyright © 2019 Pearson Education Inc. All rights reserved. Terms of Use Privacy
Answer:
It's not possible to earn an A in the courseI must have a 73 or MORE to earn a B in the courseStep-by-step explanation:
The average obtained at the end of the course will be:
[tex]\frac{85+78+84+x}{4} = av[/tex]
Where x is the grade obtained in the final examination and av is the final average. To obtain an A, av has to be at least 90, av≥90, and to obtain an B, av has to be at least 80, av≥80
Is an A in the course possible?So, if we get 100 on the final average:
x = 100,
av = (85+78+84+100)/4 = 86,75 and 86,75∠90.
Answer: No, the higher grade obtained would be 86,75.
What grade you must have in the final to earn a B in the courseTo earn a B, av≥80:
[tex]av= \frac{85+78+84+x}{4 \ }\geq 80\\ \frac{247+x}{4}\geq80 \\247+x\geq 80*4\\ x\geq 320-247\\ x\geq 73[/tex]
Answer: I must have a 73 or MORE to earn a B in the course
Write an equation of an hyperbola whose vertices are
(0,0)and(16,0), and whose foci are (18,0) and (-2,0).
Answer:
[tex]\frac{(x-8)^2}{8^2}-\frac{(y-0)^2}{6^2}=1[/tex]
Step-by-step explanation:
∵ The equation of a hyperbola along x-axis is,
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]
Where,
(h, k) is the center,
a = distance of vertex from the center,
b² = c² - a² ( c = distance of focus from the center ),
Here,
vertices are (0,0) and (16,0), ( i.e. hyperbola is along the x-axis )
So, the center of the hyperbola = midpoint of the vertices (0,0) and (16,0)
[tex]=(\frac{0+16}{2}, \frac{0+0}{2})[/tex]
= (8,0)
Thus, the distance of the vertex from the center, a = 8 unit
Now, foci are (18,0) and (-2,0).
Also, the distance of the focus from the center, c = 18 - 8 = 10 units,
[tex]\implies b^2=10^2-8^2=100-64=36\implies b = 6[/tex]
( Note : b ≠ -6 because distance can not be negative )
Hence, the equation of the required hyperbola would be,
[tex]\frac{(x-8)^2}{8^2}-\frac{(y-0)^2}{6^2}=1[/tex]
Let C(x) represent the cost of producing x items and p(x) be the sale price per item if x items are sold. The profit P(x) of selling x items is P(x)=xp(x)-C(x) (revenue minus costs). The average profit per item when x items are sold is P(x)/(x) and the marginal profit is dP/dx. The marginal profit approximates the profit obtained by selling one more item given that x items have already been sold. Consider the following cost functions C and price function p.C(x)=-0.02x^2+40x+80, p(x)=100, a=500a) what is the profit function P.P(x)=?b) find the average profit function and marginal profit function.average profit function: P(x)/(x)=?marginal profit function: dP/dx=?c
Answer:
(a) Profit function P(x) = 0.02x^2+60x-80
(b) Average profit P(x)/x = P/x = 0.02x+60-80/x
Marginal profit dP/dx = 0.04x+60
Step-by-step explanation:
Cost function: C(x) = -0.02x^2+40x+80
Price function: p(x) = 100
(a) The profit function P(x) = x*p(x)-C(x) can be expressed as:
[tex]P=x*p-C\\P=x*100-(-0.02x^{2} +40x+80)\\P=0.02x^{2}+60x-80[/tex]
(b)Average profit function: P(x)/x
[tex]P/x=(0.02x^{2}+60x-80)/x\\P/x = 0.02x+60-80/x[/tex]
Marginal profit function: dP/dx
[tex]P=0.02x^{2}+60x-80\\dP/dx=0.02*2*x+60+0\\dP/dx=0.04x+60[/tex]
Final answer:
The problem involves calculating the profit, average profit per item, and marginal profit for selling x items based on a given cost and price function. By subtracting the cost function from the revenue, we obtain the profit function P(x) = -0.02x² + 60x + 80. The average profit and marginal profit functions further analyze profitability.
Explanation:
To solve the problem given, we need to start by finding the profit function P(x), which is obtained by subtracting the cost function C(x) from the revenue function, where the revenue is the sale price per item times the number of items sold (xp(x)). Given C(x) = -0.02x² + 40x + 80 and p(x) = 100, the profit function can be determined.
Next, the average profit function is found by dividing the profit function by x, and the marginal profit function, dP/dx, is the derivative of the profit function with respect to x, which provides an approximation of the profit gained by selling one more item after x items have been sold.
Profit Function
Substituting p(x) = 100 into P(x) = xp(x) - C(x), we obtain:
P(x) = x(100) - (-0.02x² + 40x + 80)
P(x) = -0.02x² + 60x + 80
Average Profit Function
The average profit per item for x items sold is:
P(x)/x = (-0.02x² + 60x + 80) / x
what is the purpose of proof in mathematics?
The mathematical proofs are useful to show that a mathematical statement is true. Generally a mathematical proof use other statements like theorems, or axioms. Also mathematical proofs are useful to know if the development of a theoretical process in other areas like physics is well done. Other thing that is useful of the proofs in mathematics is that it use a formal language with symbols that minimize the ambiguity and make it universal.
What is the yarn number* of a 60 grain/yd sliver in cotton hanks/lb?
Answer:
0.138 hanks/lb
Step-by-step explanation:
Given:
Silver = 60 grain/yd
Now,
1 hank = 840 yd
or
1 yd = [tex]\frac{\textup{1}}{\textup{840}}[/tex] hank
And,
1 lb = 7000 grain.
or
1 grain = [tex]\frac{\textup{1}}{\textup{7000}}[/tex] lb
Thus,
60 grain/yd = [tex]\frac{60\times\frac{1}{7000}}{1\times\frac{1}{840}}[/tex] lb/hanks
or
60 grain/yd = 7.2 lb/ hanks
or
[tex]\frac{\textup{1}}{\textup{7.2}}[/tex] hanks/lb
or
0.138 hanks/lb
There are 143076 books in the library. The 3rd grade class walked over and each kid checked out 2 books. There are 21 students in the class. How many books are in left in the library after the students return to their classroom?
Answer:
The answer is: There are only 143034 books left after the students return to their classroom
Step-by-step explanation:
Lets call X= Number of books left in the library after the students return to their classroom.
B=Total of Books in the library
S=Total of books taken by the 3rd grade students
If each student of the 3rd grage class take 2 books, the total of book taken by the 3rd grade students would be:
S=2 books per student * Total of students of the 3rdgrade
S=2*21=42
Then we must substract Total of Books in the library minus Total of books taken by the 3rd grade students. That would be:
X=B-S
X=143076-42
X=143034
I hope that this answer will help you
If angle 1 has a measure of 56° and angle 2 has a measure of 124°, the two angles are complementary.
Question 1 options:
True
False
Answer:
False.
Step-by-step explanation:
Two angles are complementary when added up, they give a result of 90°.
So, to this question to be true we have to do:
Angle 1 + Angle 2 = 90
But if we resolve 56° + 124° = 180, so this means that this question is false, as the addition of both angles doesn't have a result of 90°.
Suppose an experiment has 3 stages: A, B, and C. If stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes. how many outcomes does the entire experiment have?
Answer: 72
Step-by-step explanation:
Given : An experiment has 3 stages: A, B, and C.
If stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes.
Then, by using the fundamental principle of counting (Total outcomes= product of all outcomes of each stage ) , we have
The number of outcomes the entire experiment have:-
[tex]6\times4\times3=72[/tex]
Hence, the number of outcomes the entire experiment =72
Final answer:
To find the total number of outcomes for an experiment with 3 stages having 6, 4, and 3 outcomes respectively, you multiply the outcomes of each stage together, resulting in 6 × 4 × 3 = 72 possible outcomes.
Explanation:
To determine how many outcomes the entire experiment has when an experiment has stages A, B, and C with different numbers of outcomes, you multiply the number of outcomes for each stage. Stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes. Therefore, the total number of outcomes for the entire experiment is calculated as follows:
Stage A outcomes: 6
Stage B outcomes: 4
Stage C outcomes: 3
Multiply the outcomes of each stage:
Total outcomes = 6 (Stage A) × 4 (Stage B) × 3 (Stage C) = 72 possible outcomes.
This is similar to how the sample space is determined in other contexts, such as flipping a coin and rolling a die, where you would also multiply the number of outcomes to get the size of the sample space.
Solve each formula for the given variable. State the restrictions, if any, for the formula
obtained to be meaningful.
F = ma, for a
d = [tex]\frac{m}{v}[/tex], for v
A = P + Prt, for t
A = [tex]\frac{1}{2}[/tex]h a + b, for h
P = 2(L +W ), for W
m = [tex]\frac{x+y}{2}[/tex], for y
3x + 2y = 8,for y
a = [tex]\frac{v- u}{t}[/tex], for t
It's the even numbers on this pdf:
Answer:[tex]a = \frac{F}{m}[/tex]
[tex]v = \frac{m}{d}[/tex]
[tex]t = \frac{A - P}{Pr}[/tex]
[tex]h = \frac{2A}{a + b}[/tex]
[tex]W = \frac{P - 2L}{2}[/tex]
[tex]y = 2m - x[/tex]
[tex]y = -1.5x + 4[/tex]
[tex]t = \frac{v - u}{a}[/tex]
Step-by-step explanation:
[tex]F = ma[/tex]
[tex]\frac{F}{m} = \frac{ma}{m}[/tex]
[tex]\frac{F}{m} = a[/tex]
[tex]d = \frac{m}{v}[/tex]
[tex]vd = m[/tex]
[tex]\frac{vd}{d} = \frac{m}{d}[/tex]
[tex]v = \frac{m}{d}[/tex]
[tex]A = P + Prt[/tex]
[tex]A - P = Prt[/tex]
[tex]\frac{A - P}{Pr} = \frac{Prt}{Pr}[/tex]
[tex]\frac{A - P}{Pr} = t[/tex]
[tex]A = \frac{1}{2}h(a + b)[/tex]
[tex]2A = h(a + b)[/tex]
[tex]\frac{2A}{a + b} = \frac{h(a + b)}{a + b}[/tex]
[tex]\frac{2A}{a + b} = h[/tex]
[tex]P = 2(L + W)[/tex]
[tex]P = 2(L) + 2(W)[/tex]
[tex]P = 2L + 2W[/tex]
[tex]P - 2L = 2W[/tex]
[tex]\frac{P - 2L}{2} = \frac{2W}{2}[/tex]
[tex]\frac{P - 2L}{2} = W[/tex]
[tex]m = \frac{x + y}{2}\\2m = x + y\\2m - x = y[/tex]
[tex]3x + 2y = 8\\2y = -3x + 8\\\frac{2y}{2} = \frac{-3x + 8}{2}\\y = -1.5x + 4[/tex]
[tex]a = \frac{v - u}{t}\\at = v - u\\\frac{at}{a} = \frac{v - u}{a}\\t = \frac{v - u}{a}[/tex]