we apply 35% of a drug at the morning and 25% of the same drug at the afternoon. if in the evening 28 mL of the drug is left. how many milliliters are we applying during the whole day?

Answer:

[tex]Area=\frac{7}{12}[/tex]

Step-by-step explanation:

[tex]Area=\int\limits^a_b {f(x)} \, dx =\int\limits^4_3 {\frac{7}{x^{2}} } \, dx =-7*\frac{1}{x}=-7(1/4-1/3)=\frac{7}{12}[/tex]

Answer:

There is a period of 3 consecutive days in which the website was hits at least 60 times.

Step-by-step explanation:

A web site was hit 300 times over a period of 15 days.

To solve this question we will use the Pigeonhole Principle.

Here, n = 300 and k = 5

We will find [tex]\frac{n}{k}[/tex] to get that there is a hole with at least [tex]\frac{300}{5}=60[/tex] pigeons.

Hence, there is a period of 3 consecutive days in which the website was hits at least 60 times.

Answer:

12,000.

Step-by-step explanation:

Let x be the number of people that were enrolled last year.

We have been given that enrollment at ELAC decreased by 5%, or 600 people, the year. We are asked to find the number of people that were enrolled last year.

We can set as equation such that 5% of x equals 600.

[tex]\frac{5}{100}\cdot x=600[/tex]

[tex]0.05x=600[/tex]

[tex]\frac{0.05x}{0.05}=\frac{600}{0.05}[/tex]

[tex]x=12,000[/tex]

Therefore, 12,000 people were enrolled last year.

Answer with Step-by-step explanation:

We are given that a number 8881 is not a prime number

We have to prove that it has given a prime  factor that is at most 89.

In order to prove that given number has highest prime factor is 89 we will find the prime factorization of given number.

[tex]8881=83\times 107[/tex]

Therefore, 8881 is not  a prime number and it has two factors 83 and 107.

83 is  a prime factor of 8881 which is less than 89.We have to find  a prime factor of 8881 which is atmost 89.

Therefore, 83 is that prime factor .

Hence, 8881 has a prime factor that is at most 89.

Answer: The cube root of 10 is 2.1544 using an Xo value of -0.003723

Step-by-step explanation: The Newton-Raphson is a root finding method and its formula is NR: X=Xo-(f(x)/f'(x). Once you have the equation you also need to find the derivative of that equation before applying the formula. Since the problem stated that X =10, the method was applied to find the best root in order to find the cube root of 10 up to 5 significant figures. The best method is to use a software like Excel that helps you calculate those iterations faster. The root finding for this example was -0.003723.

The probability that no passenger is in their assigned seat is 0.6321.

Given data:

There are N passengers in a plane with N assigned seats (N is a positive integer), but after boarding, the passengers take the seats randomly.

The probability that no passenger is in their assigned seat is often referred to as the "surprising" or "paradoxical" result.

The problem is analyzed for a specific case with N = 3 passengers.

Passenger 1 sits in Seat 1: In this case, the remaining two passengers have a 1/2 chance of sitting in their assigned seats.

Passenger 1 sits in Seat 2: Again, the remaining two passengers have a 1/2 chance of sitting in their assigned seats.

Passenger 1 sits in Seat 3: In this case, the remaining two passengers will definitely sit in their assigned seats.

Now, calculate the probability for each scenario and find the overall probability that no passenger is in their assigned seat:

Scenario 1: Probability = 1/3 * 1/2 = 1/6

Scenario 2: Probability = 1/3 * 1/2 = 1/6

Scenario 3: Probability = 1/3

Overall Probability = Probability of Scenario 1 + Probability of Scenario 2 + Probability of Scenario 3

= 1/6 + 1/6 + 1/3

= 1/2

Thus, for N = 3, the probability that no passenger is in their assigned seat is 1/2.

Now, consider the case when N → ∞ (approaching infinity).

In this scenario, the probability can be calculated as follows:

Probability = 1 - 1/e

Hence, as N approaches infinity, the probability that no passenger is in their assigned seat approaches 1 - 1/e, or approximately 0.6321.

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Final answer:

The derangement problem questions the probability that no passenger sits in their assigned seat, which asymptotically approaches 1/e (approximately 0.3679) as the number of seats and passengers approaches infinity.

Explanation:

The student's question is about calculating the probability that no passenger is seated in their assigned seat on a plane where passengers sit randomly, especially when the number of passengers (and seats) approaches infinity. This is known as the derangement problem or a problem of calculating permutations where no element appears in its original position. A specific case of permutations without fixed points is termed as a derangement, and the probability of a derangement occurring in a permutation of N elements approaches 1/e where e is Euler's number, approximately equal to 2.71828. Through an approximation known as the asymptotic probability, when N → [infinity], this probability approaches 1/e, which is approximately 0.3679 or 36.79%.

Answer:

466mEq

Step-by-step explanation:

First, we need to know the concentration of NaCl in a normal saline solution, this is by definition 0.9%, meaning we have 0.9g of NaCl per 100ml of solution, we want to know how much NaCl we have in 3L (3000ml):

[tex]3000ml*\frac{0.9g}{100ml}=27g=27000mg[/tex]

So, we have 27000mg in 3L of normal saline solution.

Now, acording to our milliequivalent (mEq) equation ([tex]mEq=\frac{mg}{pE}[/tex]) where pE is de molecular mass of NaCl divided by their charges, in this case 1:

[tex]pE= \frac{23+35}{1}=\frac{58}{1} = 58[/tex]

Finally we substitute in the mEq formula:

[tex]mEq=\frac{mg}{pE}=\frac{27000}{58}=466mEq[/tex]

I hope you find this information useful! Good luck!

Answer:

The diagonals of parallelogram bisect each other.

Step-by-step explanation:

Consider the provided statement.

The parallelogram has few properties.

1: The opposite side of parallelogram are parallel by the definition of parallelogram.

2: The opposite sides and angles of a parallelogram are congruent.

3: The consecutive angles of a parallelogram are supplementary.

4: The diagonals of parallelogram bisect each other.

From the above properties of a parallelogram we can say that:

The diagonals of parallelogram bisect each other.

The diagonals of a parallelogram always bisect each other, dividing each other into two equal parts. However, unless the parallelogram is a rectangle or square, the diagonals themselves are not necessarily of equal length.

In a parallelogram, the diagonals bisect each other, meaning they intersect and divide each other into two equal parts. So, if you have a parallelogram ABCD, its diagonals AC and BD will intersect at a point E, making the two parts of each diagonal (AE and EC, BE and ED) equal in length. In other words, AE equals EC and BE equals ED. However, unless the parallelogram is a special case, like a rectangle or a square, the diagonals themselves are not of equal length. Hence, the diagonals of a parallelogram are always bisecting each other but are not necessarily equal in length.

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Answer:  Option 'c' is correct.

Step-by-step explanation:

Since we have given that

The salary earned for a month stored in = A35

Rate of tax reduces by = 22%

We need to remove the % sign by dividing 22 by 100 and it becomes 0.22.

So, the dollar amount of taxes is given by

[tex]Taxes=A35\times \dfrac{22}{100}\\\\Taxes=A35\times 0.22[/tex]

Hence, Option 'c' is correct.

Answer: y = v₀tgθx - gx²/2v₀²cos²θ

a = v₀tgθ

b = -g/2v₀²cos²θ

Step-by-step explanation:

x = v₀ₓt

y = v₀y.t - g.t²/2

x = v₀.cosθt → t = x/v₀.cosθ

y = v₀y.t - g.t²/2

v₀y = v₀.senθ

y = v₀senθ.x/v₀cosθ - g/2.(x/v₀cosθ)²

y = v₀.tgθ.x - gx²/2v₀²cos²θ

a = v₀tgθ → constant because v₀ and θ do not change

b = - g/2v₀²cos²θ → constant because v₀, g and θ do not change

Final answer:

The trajectory of a projectile is shown to be parabolic by substituting time from the x-direction motion equation into the y-direction motion equation and rearranging, yielding a parabolic form y = ax + bx², with constants determined by initial velocity and gravity.

Explanation:

To prove that the trajectory of a projectile is parabolic, we start with the equations of motion in the x and y directions for a projectile that starts at the origin. For the x-direction, we have x = v_0x t, where v_0x is the initial velocity component in the x-direction and t is the time.

In the y-direction, the equation is y = v_0y t - (1/2) g t², where v_0y is the initial velocity component in the y-direction and g is the acceleration due to gravity.

To find t from the x equation: t = x / v_0x. Substituting this into the y equation yields: y = v_0y (x / v_0x) - (1/2) g (x / v_0x)^2.

Now, simplifying we get: y = (v_0y / v_0x) x - (g/2 v_0x^2) x^2, which is of the form y = ax + bx². Here, a = v_0y/v_0x and b = -g/2 v_0x²are constants depending on the initial velocity components and acceleration due to gravity. The equation y = ax + bx² represents a parabolic path, confirming that the projectile's trajectory is indeed a parabola.

Answer:

9.3125 hours

Step-by-step explanation:

Given:

Number of components consisting in a product = 25

Standard time for work cycle = 7.45 minutes

or

standard time for work cycle = [tex]\frac{\textup{7.45}}{\textup{60}}[/tex] hours

Number of units completed = 75

Now,

The number of standard hours

= Number of units completed × standard time for the work cycle

=  [tex]75\times\frac{\textup{7.45}}{\textup{60}}[/tex] hours

or

The number of standard hours = 9.3125 hours

Answer:

The answer is 9.3125 hours :)

Step-by-step explanation:

Answer:

There was 2800 tons of sand in the mound initially.

Step-by-step explanation:

Let there be m tons of sand on the beach initially.

Throughout the day, 1200 tons of sand is added to the mound. So, total sand becomes = [tex]m+1200[/tex]

Two dump trucks come in and take 800 tons of sand each from the mound.

Means they took [tex]800\times2=1600[/tex] tons of sand

At the end of the day, the mound has 2,400 tons of sand.

This can be modeled as:

[tex]m+1200-1600=2400[/tex]

Solving for m:

[tex]m-400=2400[/tex]

[tex]=> m=2400+400[/tex]

m = 2800

Hence, there was 2800 tons of sand in the mound initially.

The initial amount of sand at the beach was 4000 tons.

To solve this problem, we can set up an equation based on the given information. Let's assume that the initial amount of sand at the beach is m tons. Throughout the day, 1200 tons of sand is added, so the total amount of sand becomes m + 1200 tons. Two dump trucks take 800 tons of sand each, so the amount of sand remaining is (m + 1200) - (2 * 800) tons. At the end of the day, the mound has 2400 tons of sand, so we can set up the equation (m + 1200) - (2 * 800) = 2400.

Simplifying the equation, we have m - 1600 = 2400, which can be further simplified to m = 4000.

Therefore, the initial amount of sand at the beach was 4000 tons.

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Recall that for [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n[/tex]

Replace [tex]x[/tex] with [tex]-9x^2[/tex] and we get

[tex]\displaystyle\frac1{1-(-9x^2)}=\sum_{n\ge0}(-9x^2)^n=\sum_{n\ge0}(-9)^nx^{2n}[/tex]

Lastly, multiply this by [tex]x^3[/tex], so that

[tex]\boxed{f(x)=\displaystyle\sum_{n\ge0}(-9)^nx^{2n+3}}[/tex]

To prove that V is a vector space we must prove that the sum define on it satisfy conmutativiy, asociativity and existence of the neutral element and inverses. Also, the scalar multiplication define on V must satisfy distributivity propertie with respect to the sum and viceversa, and an asosiativity too in the sense that [tex]x(y\cdot v)= (xy)\cdot v[/tex] for [tex]x,y\in \mathbb{R}[/tex] and [tex]v\in V[/tex]. One can prove with this that the neutral element for the sum is unique. But with your operations you  have two neutral elements for [tex](1;2)[/tex]

[tex](1;2)+(-1;3)=(0;0)[/tex]

and

[tex](1;5)+(-1;11)=(0;0)[/tex]

So, you dont have a vector space.

Final answer:

The set V, with its defined addition and scalar multiplication operations, does not fulfill essential vector space properties such as the existence of an additive identity, presence of additive inverses, and correct scalar multiplication effects on components. Therefore, V is not a vector space.

Explanation:

To determine if a set V, defined with specific addition and scalar multiplication operations, is a vector space, it must satisfy several properties commonly defined in linear algebra.

For V to be considered a vector space, the addition operation must be associative and commutative, there must be an additive identity (zero vector), each vector must have an additive inverse, scalar multiplication must be associative, there must be a multiplicative identity (1), and both operations must distribute over vector addition and scalar addition.

The defined operations on V are (x₁; y₁) + (x₂; y₂) = (x₁ + x₂; 0) for vector addition and c(x; y) = (cx; 0) for scalar multiplication. These operations fail to satisfy several vector space properties, including:

Due to these inadequacies, V does not meet the criteria for a vector space under the provided operations.

Answer:

F(x)=[tex]\frac{3^x}{ln(3)}[/tex]+7cosh(x)+C

Step-by-step explanation:

The function is f(x)=3ˣ+7sinh(x), so we can define it as f(x)=g(x)+h(x) where g(x)=3ˣ and h(x)=7sinh(x).

Now we have to find the most general antiderivative of the function this means that we have to calculate [tex]\int\ {f(x)} \, dx[/tex] wich is the same as [tex]\int\ {(g+h)(x)} \, dx[/tex]

The sum rule in integration states that the integral of a sum of two functions is equal to the sum of their integrals. Then,

[tex]\int\ {(g+h)(x)} \, dx[/tex] = [tex]\int\ {g(x)} \, dx + \int\ {h(x)} \, dx[/tex]

1- [tex]\int\ {g(x)} \, dx =[/tex][tex]\int\ {3^x} \, dx = \frac{3^x}{ln(3)}+C[/tex] this is because of the rule for integration of exponencial functions, this rule is:

[tex]\int\ {a^x} \, dx =\frac{a^x}{ln(x)}[/tex], in this case a=3

2-[tex]\int\ {h(x)} \, dx =[/tex][tex]\int\ {7sinh(x)} \, dx =7\int\ {sinh(x)} \, dx =7cosh(x)+C[/tex] , number seven is a constant (it doesn´t depend of "x") so it "gets out" of the integral.

The result then is:

F(x)= [tex]\int\ {(h+g)(x)} \, dx=\int\ {h(x)} \, dx +\int\ {g(x)} \, dx[/tex]

[tex]\int\ {3^x} \, dx +\int\ {7sinh(x)} \, dx = \frac{3^x}{ln(3)} +7cosh(x) + C[/tex]

The letter C is added because the integrations is undefined.

Answer:

The expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Step-by-step explanation:

Consider the provided expression.

[tex](2x-1)^5[/tex]

The binomial theorem:

[tex](a+b)^{n}=\sum _{r=0}^{n}{n \choose r}a^{n-r}b^r[/tex]

Where,

[tex]{n \choose r}= ^nC_r =\frac{n!}{(n-r)!r!}[/tex]

Now by using the above formula.

[tex]\frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5\left(-1\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4\left(-1\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3\left(-1\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2\left(-1\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1\left(-1\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0\left(-1\right)^5[/tex]

[tex]2^5\cdot \:1\cdot \:1\cdot \:x^5-1\cdot \frac{2^4\cdot \:5x^4}{1!}+1\cdot \frac{2^3\cdot \:20x^3}{2!}-1\cdot \frac{2^2\cdot \:20x^2}{2!}+1\cdot \frac{5\cdot \:2x}{1!}+1\cdot \frac{\left(-1\right)^5}{\left(5-5\right)!}[/tex]

[tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Hence, the expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Answer:

d. uniform

Step-by-step explanation:

If the data is uniformly distributed i.e. it follows Uniform Probability Distributions then it can be visualized by a histogram. Then the shape of the histogram will be bell-shaped which means as the value of x is increases the value of y also increases for small values of y and it will decrease for a large value of y.

But, other types of probability distributions can't be visualized by a histogram.

Hence option (d) is correct.

Answer: n/2-3=13;n=32

Step-by-step explanation: the equation and solution correctly represents this sentence is n/2-3=13;n=32,

first we have to keep in mind the beginning of the exercise which is (three less) then we know that the 3 has a sign of subtraction(-3), when we meet this, we can know that this exercise has only two possible good answer.

Only the equations which have a (-3) inside will be good,the rest of equation has a 3 with a sum sign

N/2+3=13;n=5 ,

n/2+3=13;=20

then we can omit the past equations, and this let us this equations as possible.

n/2-3=13;n=32

n/2-3=13;n=8

after this we only need resolve the equation to get a correct result, only the equation with a correct result will be the correct answer,

then we proceed to clear n from each equation.

for this equation the result must be equal to 8 if we clear n.

n/2-3=13;n=8

n/2-3=13

we pass the three to the other side with sum sign

n/2=13 + 3

we resolve the sum

n/2=16

after we pass the 2 multiplying to the other side

n = 16  × 2

we resolve the product of the multiplication

n = 32

but this answer said the result must be equal to 8

n=8 ≠ n = 32

as this result is different, we can conclude that this is a bad answer,

the we get only one possibility

this equation

n/2-3=13;n=32

for this equation the result must be equal to 32 if we clear n.

n/2-3=13;n=8

n/2-3=13

we pass the three to the other side with sum sign

n/2=13 + 3

we resolve the sum

n/2=16

after we pass the 2 multiplying to the other side

n = 16  × 2

we resolve the product of the multiplication

n = 32

Answer:

C

Um I'm late but its not any of the others so....

Answer:

[tex]f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....[/tex]

Approximate error = 0.4426

Step-by-step explanation:

f(x)=tanx, a=0

Maclaurin series formula used is given below

[tex]f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)x^{n}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^{2}+\dfrac{f'''(0)}{3!}x^{3}+....[/tex]

f(x)=tanx

f(0)=tan0=0

[tex]f'(x)=sec^{2}x\\f'(0)=sec^{2}0=1\\f''(x)=2sec^{2}xtanx\\f''(0)=2sec^{2}0tan0=0\\f'''(x)=-4sec^{2}x+6sec^{4}x\\f'''(0)=-4sec^{2}0+6sec^{4}0=-4+6=2\\[/tex]

[tex]f''''(x)=-8(2sec^{2}xtan^{2}x+sec^{4}x)+24(4sec^{4}xtan^{2}x)+sec^{6})\\f''''(0)=-8(0+1)+24(0+1)=-8+24=16\\[/tex]

[tex]f(x)=0+x+0+\dfrac{2x^{3}}{3!}+\dfrac{16x^{4}}{4!}\\[/tex]

[tex]f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}\\[/tex]

Hence, the Taylor series for f(x)=tanx is given by

[tex]f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....[/tex]

Maclaurin series upper bound error formula used is given as

R_n(x)=|f(x)-T_n(x)|

R_3(x)=|tanx-T_3(x)|

[tex]R_3(x)=|tanx-x-\dfrac{x^{3}}{3}-\dfrac{2x^{4}}{3}|[/tex]

Plugging this value x=1

[tex]R_3(x)=|tan(1)-1-\dfrac{1}{3}-\dfrac{2}{3}|\\[/tex]

R_3(x)=|1.5574-1-0.333-0.666|

R_3(x)=|-0.4426|=0.4426

Hence, upper bound on the error approximation

tan(1)=0.4426

Answer:

Answered

Step-by-step explanation:

Online courses are those classes which are delivered entirely online. Students study via web cam,  chat rooms and smart boards. Whereas hybrid learning is a combination of both online learning and  traditional in class learning.  Remote work is working away from the work place at your own comfort and choice of location.

Answer:

The heat is produced by a 150-W light bulb that is on for 20-hours is 10200 BTU.

Step-by-step explanation:

To find : How much heat (Btu) is produced by a 150-W light bulb that is on for 20-hours?

Solution :

A 150-W light bulb is on for 20-hours.

The heat produced by bulb is given by,

[tex]H=150\times 20[/tex]

[tex]H=3000\ W-hr[/tex]

We know that,

[tex]1\ \text{W-hr}=3.4\ \text{BTU}[/tex]

Converting W-hr into BTU,

[tex]3000\ \text{W-hr}=3000\times 3.4\ \text{BTU}[/tex]

[tex]3000\ \text{W-hr}=10200\ \text{BTU}[/tex]

Therefore, The heat is produced by a 150-W light bulb that is on for 20-hours is 10200 BTU.

Answer:

int t(int n){

     if(n == 1)

          return 1;

     else

          return n*n*t(n-1);

}

Step-by-step explanation:

A recursive function is a function that calls itself.

I am going to give you an example of this algorithm in the C language of programming.

int t(int n){

     if(n == 1)

          return 1;

     else

          return n*n*t(n-1);

}

The function is named t. In the else clause, the function calls itself, so it is recursive.

Step-by-step explanation:

The statement to be proved using mathematical induction is:

We will begin the proof showing that the base case is satisfied (n=4).

[tex]5^4=625\geq 612=2^{2*4+1}+100[/tex].

Then, 1 is true for n=4.

Now we will assume that the statement holds for some arbitrary natural number [tex]n\geq 4[/tex] and prove that then, the statement holds for n+1. Observe that

[tex]2^{2(n+1)+1}+100=2^{2n+1+2}+100=4*2^{2n+1}+100\leq 4(2^{2n+1}+100)\leq 4*5^n<5^{n+1}[/tex]

With this the inductive step has been proven and then, our statement is true,

For every [tex]n\geq 4[/tex], [tex]5^n\geq 2^{2n+1}+100[/tex]

Answer:

(a) 0.8512 (b) 0.8512 (c) 0.7024 (d) 0.0110

Step-by-step explanation:

The blood glucose follows a normal distribution N(μ=85;σ=24).

For every value of X, we can calculate the z-score (equivalent for a N(0;1)) and compute the probability.

(a) P(x>60)

z = (x-μ)/σ = (60-85)/24 = -1.0417

P(x>60) = P(z>-1.0417) = 0.8512

(b) P(x<110)

z = (x-μ)/σ = (110-85)/24 = 1.0417

P(x<110) = P(z<1.0417) = 0.8512

(c) P(60<x<110) = P(x<110)-P(x<60)

P(60<x<110) = P(z<1.0417) - P(z<-1.0417)

P(60<x<110) = 0.8512 - (1-0.8512) = 0.8512 - 0.1488 = 0.7024

(d) P(x>140)

z = (x-μ)/σ = (140-85)/24 = 2.2917

P(x>140) = P(z>2.2917) = 0.0110

Final answer:

Explanation of probabilities for different blood glucose levels using mean and standard deviation.

Explanation:

Probability calculations for blood glucose levels:

(a) x is more than 60: Calculate the z-score using the formula z = (x - μ) / σ. With x = 60, μ = 85, and σ = 24, find the probability using a standard normal distribution table.

(b) x is less than 110: Use the z-score formula with x = 110, μ = 85, and σ = 24 to determine the probability.

(c) x is between 60 and 110: Find the individual probabilities for x = 60 and x = 110, then subtract the two values to get the probability in this range.

(d) x is greater than 140: Similar to the previous steps, find the z-score for x = 140 and calculate the probability.

Answer:

Ans. the amount of time needed for the sinking fund to reach $25,000 if invested $255/month at 5.8% compounded monthly (Effective monthly=0.4833%) is 80.45 months.

Step-by-step explanation:

Hi, first we need to transform that 5.8% compounded monthly into an effective monthly rate, that is as follows.

[tex]r(EffectiveMonthly)=\frac{r(Comp.Monthly)}{12} =\frac{0.058}{12} =0.00483[/tex]

That means that our effective monthly rate is 0.483%

Now, we need to solve for "n" the following formula.

[tex]FutureValue=\frac{A((1+r)^{n}-1) }{r}[/tex]

Let´s start solving

[tex]25,000=\frac{255((1.00483)^{n}-1) }{0.00483}[/tex]

[tex]\frac{25,000*0.00483}{255} =1.00483^{n} -1\\[/tex]

[tex]0.47352941=1.00483^{n} -1[/tex]

[tex]1+0.47352941=1.00483^{n[/tex]

[tex]1.47352941=1.00483^{n}[/tex]

[tex]Ln(1.47352941)=n*Ln(1.00483)[/tex]

[tex]\frac{Ln(1.47352941)}{Ln(1.00483)} =n=80.45[/tex]

This means that it will take 80.45 months to reach $25,000 with an annuity of $255 at a rate of 5.8% compounded monthly (0.4833% effective monthly).

Best of luck.

Answer:

Ac = 10100100011

[tex]A \cup B = 11111111110[/tex]

[tex]A \cap B = 00010011000[/tex]

[tex]A - B = 100100111010[/tex]

Step-by-step explanation:

All these operations are bitwise operations.

Ac is the complement of a. So where we have a bit 0, the complement is 1. Where we have a bit 1, the complement is 0. So

A = 01011011100

Ac = 10100100011

The second operation is the union between A and B. This is bitwise(bit 1 of A with bit 1 of B, bit 2 of A with bit 2 of B,...). The union operation is only 0 when it is between two zeros. So:

A = 01011011100

B = 10110111010

[tex]A \cup B = 11111111110[/tex]

The third operation is the intersection between A and B. Again, bitwise. The intersection is only 1 when it is between two bits that are 1. So

A = 01011110100

B = 10110111010

[tex]A \cap B = 00010011000[/tex]

The last operation is the bitwise subtraction between A and B. We start from the least significant bit(the last one). And we have to take care of the borrow operator also, similarly to a decimal subtraction.

We can only borrow from a previous bit 1, and this bit is set to 0

0-0 is 0 with no borrow

1-0 is 1 with no borrow

1 with borrow - 0 is 1 with borrow

1 with borrow - 1 is 1 with no borrow

0-1 is 1 with no borrow

0 with borrow -1 is 0 with no borrow

1-1 is 0 with no borrow

If we arrive at the first bit(the most significant) with a borrow, we must add a 1 at the front of the answer. So

A = 01011110100

B = 10110111010

[tex]A - B = 100100111010[/tex]

Final answer:

To find the length of AC in the right-angled isosceles triangle △ADC with altitude CD and AD equals to BC, we use the Pythagorean theorem, substituting known lengths to solve for AC, which is √11.25 cm.

Explanation:

The student is asking to find the length of side AC in a triangle △ABC where CD is an altitude and AD is equal to BC, with given lengths AB = 3 cm and CD = 3 cm.

To solve for AC, we can use the Pythagorean theorem which states that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Since CD is an altitude and AD = BC, △ADC is a right-angled isosceles triangle.

Using the Pythagorean theorem in △ADC, we have:

AD2 + CD2 = AC2

Since AD = BC and CD = 3 cm, let's assume AD = x cm. Then,

x2 + 32 = AC2

x2 + 9 = AC2

But, AB = 3 cm, and AB = AD + DB = x + x = 2x,

Therefore, x = AB/2 = 1.5 cm. Substituting this value back into the previous equation:

(1.5)2 + 9 = AC2 = 2.25 + 9 = 11.25

AC = √11.25 cm

Therefore, the length of AC is √11.25 cm.

The total volume of the bird feeder is 47.1 cubic inches.

The total volume of the bird feeder.

Given:

So, the total volume of the bird feeder is 47.1 cubic inches.

The total distance from A to B is 5 ( -3 to 2 = 5).

Using the ratio 1/5, split the distance in to 1/5th's, point K would be at -2.

Answer:

A to B is 5 ( -3 to 2 = 5).//!!//Then you are gonna be Using the ratio 1/5, split the distance in to 1/5th's, point K would be at -2.

Answer: time = 55.48 years

Explanation:

Given:

Principal amount = $10000

Interest rate = 2% p.a

Amount = $30000

We can evaluate the time taken using the following formula:

[tex]Amount=Principal(1+\frac{r}{100})^{t}[/tex]

[tex]30000 = 10000\times(1+\frac{2}{100})^{t}[/tex]

Solving the above equation, we get

[tex]1.02^{t} = 3[/tex]

Now taking log on both sides, we get;

[tex]t\ log(1.02) = log(3)[/tex]

time = 55.48 years