A piece of cardboard has two circles punched out of it. What is the approximate area of the remaining cardboard? Use 3.14 for pie and round to the nearest whole number.
227 cm2
246 cm2
258 cm2
276 cm2
Answers
we know that
[the approximate area of the remaining cardboard]=(area of rectangle) - 2(area of circle)
area of rectangle=14 cm* 18 cm------> 252 cm²
area of a circle=pi*r²
r=2/1----> 1 cm
area=pi*1²-----> 3.14 cm²
[the approximate area of the remaining cardboard]=252- 2*3.14
=245.72-------> 246 cm²
the answer is
246 cm²
Related Questions
Runners in a long distance race start out going 5 kilometers south and then head west for the remainder of the race. The finish line is 13 kilometers from the starting line. How far did the runners travel?
Answers
5 km +12 km = 17 km
_____
By the Pythagorean theorem, ...
(straight-line distance)² = (distance south)² +(distance west)²
In km, this is
13² = 5² +(distance west)²
169 -25 = (distance west)²
√144 = 12 = distance west
6.
Valdez Construction signed a note with a payment of $5,200 per quarter for 5 years.
Find the amount they must set aside today to satisfy this capital requirement in an account earning 8% compounded quarterly.
$30,506.32
$126,346.32
$51,054.38
$85,027.44
Answers
A = P [1-(1+R/4)^4n]/(R/4) = 5200 [1-(1+0.08/4)^4*5]/(0.08/4) = $85,027.45
Therefore, Valdez construction should set aside $85,027.44 now to satisfy the capital requirement.
Which pair of triangles can be proven congruent by using the HL theorem?
Answers
Explanation:
The HL theorem is the hypotenuse leg theorem.
The HL theorem is referred to the congruency of right triangles.
This theorem states that two right triangles that have a congruent hypotenuse and a corresponding, congruent leg are congruent triangles.
Figure A shows the congruency of two pairs of legs. So, this is not the answer.
Figure B shows the congruency of one pair of legs and one pair of angles. So, this is not the answer.
Figure C. shows the congruency of the two hypotenuses and one pair of legs. So, this is the righ answer.
Suppose you want to transform the graph of the function y=tan(x+pi/4)-1 into the graph of the function y=-tan(x+pi/2)+1
Answers
In the given equation
+1 will shift the function upward by 1 unit
minus(-) sign will reflect the graph across x_axis
see the attached diagram of -tan(x+π/2) +1
Answer:
A) Reflect the graph of the first function across the x-axis, translate it pi/4 units to the left, and translate it 2 units up
what is 12.81 repeated rounded to the nearest hundredth
Answers
Let me know if you have any further questions.
:)
estimate the difference between 9,030 and 738
Answers
9,030 ⇒ 9,000
738 ⇒ 700
9,000 - 700 = 8,300
simplify 3.2-5.1n-3n+5
Answers
-8.1n+8.2
3.2+5= 8.2
-5.1n-3n= -8.1n
Now you combine the two and get an answer of -8.1n-8.2
will give braniliest....... Mom put plums and apples onto a plate. The ratio of the number of plums to the number of apples was 3:2. How many fruit did mom put on the plate, if after Ed took 6 there number of plums on the plate became the same as the number of apples?
Answers
So she would have put 30 fruits (18 plums and 12 apples) on the plate
Hope this helps!
Solve the equation <(a-5)-5=3
Answers
a - 5 - 5 = 3
Calculate the difference between -5 and 5.
a - 10 = 3
Now move the constant to the right side of the equation and change its sign.
a = 3 + 10
Add the numbers on the right side together to get your final answer.
a = 13
Let me know if you have any further questions.
:)
Construct a 90% confidence interval for the population mean, µ. assume the population has a normal distribution. a sample of 15 randomly selected math majors has a grade point average of 2.86 with a standard deviation of 0.78. round to the nearest hundredth
Answers
The 90% confidence interval for the population mean [tex]\( \mu \)[/tex] is approximately (2.50, 3.21) .
To construct a 90% confidence interval for the population mean [tex]\( \mu \)[/tex], we can use the formula:
[tex]\[ \text{Confidence Interval} = \bar{x} \pm \left( \text{Critical Value} \times \frac{s}{\sqrt{n}} \right) \][/tex]
where:
- [tex]\( \bar{x} \)[/tex] is the sample mean,
- s is the sample standard deviation,
- n is the sample size, and
- the critical value corresponds to the desired confidence level and degrees of freedom.
Given:
- Sample mean [tex]\( \bar{x} = 2.86 \),[/tex]
- Sample standard deviation s=0.78
- Sample size n=15
- Confidence level = 90%.
First, we need to find the critical value corresponding to a 90% confidence level and 14 degrees of freedom (since ( n - 1 = 15 - 1 = 14 )). We can find this value using a t-distribution table or a statistical calculator. For a 90% confidence level and 14 degrees of freedom, the critical value is approximately 1.7613.
Now, let's calculate the confidence interval:
[tex]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times \frac{0.78}{\sqrt{15}} \right) \]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times \frac{0.78}{\sqrt{15}} \right) \]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times \frac{0.78}{3.87298} \right) \]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times 0.20172 \right) \]\[ \text{Confidence Interval} = 2.86 \pm 0.35587 \][/tex]
Lower Limit:
[tex]\[ 2.86 - 0.35587 \approx 2.5041 \][/tex]
Upper Limit:
[tex]\[ 2.86 + 0.35587 \approx 3.2141 \][/tex]
a dress was reduced from $100 to $85. express the discount as a % of the original price
Answers
$15/$100 = 15/100 = 15%
_____
It can be helpful to think of "%" as another way to write "/100".
Please Help!
Khalid has a game board as shown below, which is a square with 20-cm sides. The area of the largest circle is 320 square centimeters.
What is the probability of scoring 1, 3, or 5 points with one randomly thrown dart?
A. 1/2
B. 5/8
C. 3/4
D. 4/5
Answers
Answer:
Option D. 4/5
Step-by-step explanation:
we know that
The probability of scoring 1, 3, or 5 points with one randomly thrown dart is equal to divide the area of the largest circle by the area of the square game board
step 1
Find the area of the square game board
[tex]A=b^{2}[/tex]
we have
[tex]b=20\ cm[/tex]
substitute
[tex]A=20^{2}[/tex]
[tex]A=400\ cm^{2}[/tex]
step 2
Find the probability
[tex]P=320/400[/tex]
[tex]P=0.8=8/10=4/5[/tex]
What is the first step in solving ln(x − 1) = ln6 − lnx for x?
Answers
Answer:
[tex]ln(x - 1) = ln(\frac{6}{x})[/tex]
Step-by-step explanation:
[tex]ln(x - 1) = ln6 - lnx[/tex]
To solve for x we need to simplify the ln
To simplify logarithmic function we use log property
[tex]ln(a) - ln(b) = ln(\frac{a}{b})[/tex]
we apply the same property on the right hand side of the given equation
[tex]ln(x - 1) = ln6 - lnx[/tex]
[tex]ln(6) - ln(x) = ln(\frac{6}{x})[/tex]
[tex]ln(x - 1) = ln(\frac{6}{x})[/tex]
This is the first step in solving the given equation
PLZ HELP ASAP FEATURES OF A CIRCLE FROM ITS EXPANDED EQUATION
Answers
[tex] x^{2} +y^{2} +14x-10y+65=0 \\ \\ x^{2} +2(x)(7)+ y^{2}-2(y)(5)=-65 \\ \\ x^{2} +2(x)(7)+ (7)^{2}+ [y^{2}-2(y)(5)+ (5)^{2}]=-65+(7)^{2}+(5)^{2} \\ \\ (x+7)^{2}+(y-5)^{2}=9 [/tex]
The above equation is in standard form.
The center of the circle is (-7, 5) and its radius is 3.
Design amanda wants to make this design of circles inside an equilateral triangle.
a. what is the radius of the large circle to the nearest hundredth of an inch?
b. what are the radii of the smaller circles to the nearest hundredth of an inc
Answers
the answer in the attached figure
Answer:
Using theorem
AE=8.66
Thus radius of large circle is one third of equilateral triangle altitude.
Radius of larger circle=2.9 inch
And radius of inner circle will be 0.96 inch
hey can you please help me posted picture of question
Answers
The solutions are 3 and -3
Explanation:
To get the solution, we will need to isolate the x on one side of the equation as follows:
10x² - 56 = 88 - 6x²
10x² - 56 + 56 = 88 - 6x² + 56
10x² = 144 - 6x²
10x² + 6x² = 144 - 6x² + 6x²
16x² = 144
x² = 144/16
x² = 9
x = ±√9
either x = +√9 = 3
or x = -√9 = -3
Hope this helps :)
Make up an equation of the form y = kx +b, the graph of which passes through the following points: C (–19, 31) and D (1, –9)
Answers
it is possible to make up the system of two equations using the coordinates of points C and D:
31=k*(-19)+b - if to substitute the coordinates of point C;
-9=k*1+b - if to substitute the coordinates of point D.
[tex] \left \{ {{-19k+b=31} \atop {k+b=-9}} \right. \ =\ \textgreater \ \ \left \{ {{20k=-40} \atop {k+b=-9}} \right. \ =\ \textgreater \ \ \left \{ {{k=-2} \atop {b=-7}} \right. [/tex]
y=-2x-7
answer: y=-2x-7
Two friends bring hamburger meat to your cookout. One brings 2.7 pounds, and the other brings 3.54 pounds. How much hamburger meat do they bring?
Answers
They bought 6.24 pounds of hamburger meat
Explanation:
We are given that:
The first friend brought 2.7 pounds of hamburger meat
The second friend brought 3.54 pounds of hamburger meat
The total amount of hamburger meat they both brought would be simply the summation of what they brought individually.
This means that:
Total amount brought = 2.7 + 3.54
Total amount brought = 6.24 pounds of hamburger meat
Hope this helps :)
For the data set below, calculate the standard deviation to the nearest hundredth decimal place. 27 38 47 42 33 56 37 57 38 52
Answers
Final answer:
The standard deviation of the given data set is 10.03, and the value that is one standard deviation below the mean is 32.67. Calculations involve finding the mean, computing squared deviations, calculating the sum of squared deviations, finding the variance, and taking the square root of the variance.
Explanation:
The mean is calculated as: (27 + 38 + 47 + 42 + 33 + 56 + 37 + 57 + 38 + 52) ÷ 10 = 427 ÷ 10 = 42.7.
Now we calculate each deviation from the mean, square it, and sum them:
(27 - 42.7)² = 246.49
(38 - 42.7)² = 22.09
(47 - 42.7)² = 18.49
(42 - 42.7)² = 0.49
(33 - 42.7)² = 94.09
(56 - 42.7)² = 176.89
(37 - 42.7)² = 32.49
(57 - 42.7)² = 204.49
(38 - 42.7)² = 22.09
(52 - 42.7)² = 86.49
A sum of squared deviations = 904.61.
The variance is 904.61 ÷ (10-1) = 100.51.
The standard deviation is the square root of the variance,
which is √100.51 equal to approximately 10.03.
To find the value that is one standard deviation below the mean, we subtract one standard deviation from the mean: 42.7 - 10.03 = 32.67.
Therefore, the standard deviation to the nearest hundredth is 10.03, and the value that is one standard deviation below the mean is approximately 32.67.
Given that tan^2 e= 3/8 what is the value of sec e? A. +√8/3 B.+ √11/8 C. 11/8 D. 8/3
Answers
ANSWER
[tex]{ \sec}(e) = \pm \: \sqrt{\frac{11}{8} } [/tex]
EXPLANATION
We use the Pythagorean Identity,
[tex] { \sec}^{2} (e) = 1 + { \tan}^{2} (e)[/tex]
It was given that,
[tex] { \tan}^{2} (e) = \frac{3}{8} [/tex]
We substitute the values into the identity to obtain,
[tex] { \sec}^{2} (e) = 1 + \frac{3}{8} [/tex]
[tex]{ \sec}^{2} (e) = \frac{11}{8} [/tex]
We take square root of both sides to get,
[tex]{ \sec}(e) = \pm \: \sqrt{\frac{11}{8} } [/tex]
Answer:
sec e = √(11/8) ⇒ answer B
Step-by-step explanation:
* Lets revise some identities in trigonometry
# sin²x + cos²x = 1
- Divide both sides by cos²x
∴ sin²x/cos²x + cos²x/cos²x = 1/cos²x
∵ sinx/cosx = tanx
∴ sin²x/cos²x = tan²x
∵ cos²x/cos²x = 1
∵ 1/cosx = secx
∴ 1/cos²x = sec²x
* Now lets write the new identity
# tan²x + 1 = sec²x
- Let x = e
∴ tan²e + 1 = sec²e
- Substitute the value of tan²e in the identity
∵ tan²e = 3/8
∴ 3/8 + 1 = sec²e
- Change the 1 to the fraction 8/8
∴ 3/8 + 8/8 = sec²e ⇒ add the fractions
∴ 11/8 = sec²e
- Take square root for the two sides to find sec e
∴ sec e = √(11/8)
∴ The answer is B
the square in this figure has a side length of 14 inches. the radius of the quarter circle is 7 inches. what is the estimate area
Answers
The area of the shaded region is 42.14 square inches.
How the area of the shaded region is determined:
We are given a square with side lengths of 14 inches, and each corner of the square has a quarter circle with a radius of 7 inches. We need to find the area of the shaded region, which is the area of the square minus the area of the quarter circles.
To calculate the area of the figure, we need to consider both the square and the quarter circles.
1. Calculate the area of the square:
The square has a side length of 14 inches.
Area of the square = [tex]side^2[/tex] = [tex]14^2[/tex] = 196 square inches
2. Calculate the area of the quarter circles:
The quarter circle has a radius of 7 inches.
The area of a full circle is given by:
Area of the circle = [tex]\pi r^2[/tex]
Area of the quarter circle = 4/4 x [tex]\pi r^2[/tex]
= 1 x 3.14 x [tex]7^2[/tex]
= 153.86 square inches
Therefore, the area of the shaded region = Area of the square - Area of the quarter circles:
= 196 - 153.86
= 42.14 square inches
Complete Question:
The square in the figure has a side length of 14 inches. The radius of the each quarter circle is 7 inches. What is the area of the shaded region?
1. Clyde has the chance to buy a piece of old Pennsylvania Dutch pottery that he thinks he can resell for $500. If Clyde needs a 125% markup on cost, what price should he pay?
2. Orchard Supply sells lawn fertilizer at a price of $12.50 per bag. If the markup is 25% of cost, find the cost.
Answers
2.25*cost = 500
cost = 500/2.25 = 222.22
Clyde should pay $222.22 or less for the pottery.
2. cost +0.25*cost = $12.50
1.25*cost = $12.50
cost = $12.50/1.25 = $10.00
The cost to Orchard Supply of a bag of fertilizer is $10.00.
An eagle can fly at a speed of 50 mph and a starling can fly at 78 mph. How far will the starling fly in the time it takes the eagle to fly 125 miles?
PLZ HELP ASAP!!!!!!
20 POINTS!
Answers
1.
50*x=125
x=2.5
2.
2.5 * 78 = 195 miles
Answer:
195 miles
Step-by-step explanation:
50*x=125
x=2.5
2.5*78=195miles
The scatter plot shows the relationship between the number of car accidents in a month and the number of drivers attending a program on distracted driving. The equation represents the linear model for this data. y=−0.0067x+17 What does the number -0.0067 in the equation mean in this context? There were 0.67 accidents per month. The number of accidents was reduced by 0.67 per month for every additional 100 drivers in the program. The number of accidents was reduced by 0.67 per month for every additional driver in the program. The number of accidents increased by 0.67 per month for every additional 100 drivers in the program. The number of accidents was reduced by 0.67 per month every month.
Answers
Answer:
the answer is A
Use parametric equations of the ellipse, ???? 2 16 + ???? 2 9 = 1, to find the area that it encloses in the first quadrant.
Answers
Two planes are flying in opposite directions, away from each other, one with the speed of 800 km per hour and the other with the speed of 840 km per hour. How much farther from each other are the planes getting every hour?
Answers
A patient is given a 50 mg dose of medicine the medicines effectiveness decreases every hour at a constant rate of 40% what is the exponential decay function that models this scenario how much medicine will be left in the patient's system after two hours
Answers
[tex]y = a(1 - r)^{x} \\ y = 50(1 - .40) ^{x} [/tex]
x= hours past
[tex]y = 50(1 - .40)^{2} \\ y = 18[/tex]
after 2 hours, there are 18 mg of medicine left
Answer:
[tex]A=50(0.6)^x[/tex]
18 mg of medicine will be left in the patient's system after two hours.
Step-by-step explanation:
Given,
The initial quantity of the medicine, P = 50 mg,
Also, it decreases every hour at a constant rate of 40%
That is, r = 40 %,
Thus, the quantity of the medicine after x hours,
[tex]A=P(1-\frac{r}{100})^r[/tex]
[tex]=50(1-\frac{40}{100})^x[/tex]
[tex]=50(1-0.4)^x[/tex]
[tex]=50(0.6)^x[/tex]
Which is the required exponential decay function that models this scenario.
The quantity of the medicine after 2 hours,
[tex]A=50(0.6)^2=18\text{ mg}[/tex]
Joey got a 25% raise on his salary. if his original salary was 1,200, how much was it after the raise was implemented?
Answers
What is the value of x in the following equation 1/5x=5^8
A)-8 B) 8 C)40 D)-40
Answers
What is the value of x in the following equation (1/5)^x=5^8
Applying logarithms both members
log(1/5)^x=log(5^8)
x*log(1/5)=8*log(5)
x=8*log(5)/log(1/5)
we know that
log (1/5)----> log (1)-log (5)----> 0-log (5)-----> -log (5)
so
x=8*log(5)/-log (5)------> x=-8
the answer is
-8
One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. let p represent the number of female insects in a population and s the number of sterile males introduced each generation. let r be the per capita rate of production of females by females, provided their chosen mate is not sterile. then the female population is related to time t by t = p + s p[(r − 1)p − s] dp. suppose an insect population with 10,000 females grows at a rate of r = 1.2 and 400 sterile males are added. evaluate the integral to give an equation relating the female population to time. (note that the resulting equation can't be solved explicitly for p. remember to use absolute values where appropriate.)
Answers
To be clear, the given relation between
time and female population is an integral:
[tex]t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP [/tex]
[tex]t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]} } \, dP [/tex]
= [tex] \int { \frac{P+400}{P(0.2P - 400)} } \, dP [/tex]
In order to evaluate this integral, we need to write this rational function in a simpler way:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} + \frac{B}{(0.2P - 400)} [/tex]
where we need to evaluate A and B. In order to do so, let's calculate the LCD:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P - 400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)} [/tex]
the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
= P(0.2A + B) - 400A
The two sides must be equal to each other, bringing the system:
[tex] \left \{ {{0.2A + B = 1} \atop {-400A = 400}} \right. [/tex]
Which can be easily solved:
[tex] \left \{ {{B=1.2} \atop {A=-1}} \right. [/tex]
Therefore, our integral can be written as:
[tex]t = \int { \frac{P+400}{P(0.2P - 400)} } \, dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP [/tex]
= [tex]- \int { \frac{1}{P} \, dP + 1.2\int { \frac{1}{0.2P-400} } \, dP[/tex]
= [tex]- \int { \frac{1}{P} \, dP + 6\int { \frac{0.2}{0.2P-400} } \, dP[/tex]
= - ln |P| + 6 ln |0.2P - 400| + C
Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5)
Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)