Dwayne's garden is triangle-shaped with two equal sides and a third side that is 4 ft more than the length of an equal side. If the perimeter is 49 ft, how long is each side

Answers

Answer 1

A triangle perimeter = side1+side2+side3=49ft

Where side1=side2 because 2 of the sides equal

And side3=side1+4ft because side3 is 4ft more than the length of an equal side

Plug them in
49ft=side1+side1+(side1+4ft)

simplify
Subtract 4 from both sides
45ft = side1+side1+side1
45ft = 3*side1
Divide both sides by 3
15ft = side1

Side1 =side2=15ft
Side3=side1+4ft= 15+4=19ft

Related Questions

Adriana's water bottle contains 2 quarts of water she wants to add a drink to mix into it but the directions for the treatments gives a amount of water fluid ounces how many fluid ounces are in her bottle

Answers

2 quarts of water=64 fluid ounces

On a digital clock, the colon (two dots between the hour and minutes) blinks every second. In 2 hours and 20 minutes, how many times will it blink?

Answers

2·60=120
120+20=140
140·60=8400

8400

See Picture

The colon will blink 8400 times in 2 hours and 20 minutes.

How many times will the colon blink?

Since the colon blinks every second. And we know that there are 60 minutes in an hour and there are 60 seconds in a minute.

Thus, we can say:

1 minutes = 60 seconds

1 hour = 60 minutes = 3600 seconds

Note: 60 minute = (60 * 60) seconds = 3600 seconds

Thus, in 2 hours and 20 minutes, we have:

2 hours + 20 minutes = (2 * 3600) + (20 * 60)

= 7200 + 1200

= 8400 second

Therefore, the colon will blink 8400 times in 2 hours and 20 minutes.

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n a study of 250 adults, the mean heart rate was 70 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 12 beats per minute. What is the 99% confidence interval for the mean beats per minute?

Answers

We are given mean = 70, SD = 12, sample size n = 250, and for a 99% confidence interval, the two-tailed z-score is +/-2.58.
The CI bounds are calculated as:
mean +/- z*SD/sqrt(n) = 70 +/- 2.58*12/sqrt(250) = 68.04 to 71.96 beats per minute

Find the lateral area for the regular pyramid.

L. A. =

Answers

The lateral area is:
Al = (4) * (1/2) * (b) * (h)
Where,
b: base of the triangle
h: height of the triangle
Al = (4) * (1/2) * (2) * (root ((1) ^ 2 + (3) ^ 2))
Al = (4) * (root (1 + 9))
Al = 4raiz (10) units ^ 2
Answer:
Al = 4raiz (10) units ^ 2

The answer would be Al = 4raiz (10) units ^ 2Formula:
Al = (4) * (1/2) * (b) * (h)
Where, b= base of the triangle h=height of the triangle Al = (4) * (1/2) * (2) * (root ((1) ^ 2 + (3) ^ 2)) Al = (4) * (root (1 + 9)) Al = 4raiz (10) units ^ 2

Someone please help me?

Answers

(x-3)^2 + (y+5)^2 = 25

remove the ^2's from the left side:

(x-3) +(y+5) = sqrt(25)

(x-3) +(y+5) = 5

radius = 5

now solve for zero 's on the left side:

x-3 = 0 x = 3 (3-3=0)

y+5 = 0 y = -5 (-5 +5=0)

so center = (3,-5) and radius is 5

Answer is last one.

The domain of f(x)=2logx+3 is x > 3.

true.
false.

Answers

This is a problem of calculus. So, for getting the domain of this function, we nee to study each term. So, firs of all we have the function log(x). For this function it is true that the domain is:

x > 0

So, there is not other function for this problem. Thus, the answer is false. See the figure below that shows f(x)

Use the unit circle to find the value of each trigonometric function at the angle indicated

Answers

0
-1
-inf
1
0
0

repectively

Answer:

Step-by-step explanation:

We have to find the values of the given trigonometric ratios at the angle indicated. Thus,

(A) The given trigonometric function is:

[tex]cos270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]cos270^{\circ}=-(0)=0[/tex]

(B) The given trigonometric function is:

[tex]sin270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]sin270^{\circ}=-1[/tex]

[tex]tan270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]tan270^{\circ}=undefined[/tex]

(D) The given function is:

[tex]cos0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also, [tex]cos0^{\circ}=1[/tex]

(E) The given function is:

[tex]sin0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also, [tex]sin0^{\circ}=0[/tex]

(F) The given function is:

[tex]tan0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also, [tex]tan0^{\circ}=0[/tex]

A climber is standing at the top of mount kazbek, approximately 3.1 miles above sea level. the radius of the earth is 3959 miles. what is the climber's distance to the horizon? enter your answer as a decimal in the box. round only your final answer to the nearest tenth. mi

Answers

In the figure below, the radius, AB=AD=3959 mi, the climber is at position C. BC=3.1 mi. CD=x mi, is the distance from the climber to the horizon. Thus to solve for x we proceed as follows:
AC=3959+3.1=3962.1 mi
To evaluate for x we use the Pythagorean theorem, this is given by:
c^2=a^2+b^2
c=AC is the hypotenuse
a and b are the legs
plugging in the values we shall have:
3962.1^2=x^2+3959^2
x^2=3961.1^2-3959^2
x^2=24555.41
hence
x=156.7 mi

Answer: 156.7 mi

The nearest tenth, the climber's distance to the horizon is approximately 156.6 miles.

The climber's distance to the horizon can be calculated using the Pythagorean theorem, where the Earth's radius and the climber's height above sea level form a right-angled triangle.

[tex]\[ d^2 + R^2 = (R + h)^2 \][/tex]

Expanding the right side of the equation gives us:

[tex]\[ d^2 + R^2 = R^2 + 2Rh + h^2 \][/tex]

Subtracting [tex]$ R^2 $[/tex] from both sides, we get:

[tex]\[ d^2 = 2Rh + h^2 \][/tex]

Since [tex]$ h^2 $[/tex] is very small compared to [tex]$ 2Rh $[/tex] (because [tex]$ h $[/tex] is much smaller than [tex]$ R $)[/tex], we can neglect [tex]$ R^2 $[/tex] in our calculation for a more practical approximation. This leaves us with:

[tex]\[ d^2 \approx 2Rh \][/tex]

Taking the square root of both sides to solve for [tex]$ d $[/tex], we have:

[tex]\[ d \approx \sqrt{2Rh} \][/tex]

Now, plugging in the values for [tex]$ R $ and $ h $:[/tex]

[tex]\[ d \approx \sqrt{2 \times 3959 \times 3.1} \][/tex]

[tex]\[ d \approx \sqrt{24516.8} \][/tex]

[tex]\[ d \approx 156.6 \text{ miles} \][/tex]

Rounding to the nearest tenth, the climber's distance to the horizon is approximately 156.6 miles.

. (06.02)
The table below shows data for a class's mid-term and final exams:

Mid-Term Final
96 100
95 85
92 85
90 83
87 83
86 82
82 81
81 78
80 78
78 78
73 75

Which data set has the smallest IQR? (1 point)

They have the same IQR
Mid-term exams
Final exams
There is not enough information
2. (06.02)
The box plots below show student grades on the most recent exam compared to overall grades in the class:

two box plots shown. The top one is labeled Class. Minimum at 74, Q1 at 78, median at 85, Q3 at 93, maximum at 98. The bottom b
Which of the following best describes the information about the medians? (1 point)

The exam median is only 1–2 points higher than the class median.
The exam median is much higher than the class median.
The additional scores in the second quartile for the exam data make the median higher.
The narrower range for the exam data causes the median to be higher.
3. (06.02)
The box plots below show attendance at a local movie theater and high school basketball games:

two box plots shown. The top one is labeled Movies. Minimum at 60, Q1 at 65, median at 95, Q3 at 125, maximum at 150. The botto
Which of the following best describes how to measure the spread of the data? (1 point)

The IQR is a better measure of spread for movies than it is for basketball games.
The standard deviation is a better measure of spread for movies than it is for basketball games.
The IQR is the best measurement of spread for games and movies.
The standard deviation is the best measurement of spread for games and movies.
4. (06.02)
The box plots below show the average daily temperatures in April and October for a U.S. city:

two box plots shown. The top one is labeled April. Minimum at 50, Q1 at 60, median at 67, Q3 at 71, maximum at 75. The bottom b
What can you tell about the means for these two months? (1 point)

The mean for April is higher than October's mean.
There is no way of telling what the means are.
The low median for October pulls its mean below April's mean.
The high range for October pulls its mean above April's mean.
5. (06.02)
The table below shows data from a survey about the amount of time high school students spent reading and the amount of time spent watching videos each week (without reading):

Reading Video
5 1
5 4
7 7
7 10
7 12
12 15
12 15
12 18
14 21
15 26

Which response best describes outliers in these data sets? (2 points)

Neither data set has suspected outliers.
The range of data is too small to identify outliers.
Video has a suspected outlier in the 26-hour value.
Due to the narrow range of reading compared to video, the video values of 18, 21, and 26 are all possible outliers.
6. (06.02)
Male and female high school students reported how many hours they worked each week in summer jobs. The data is represented in the following box plots:

two box plots shown. The top one is labeled Males. Minimum at 0, Q1 at 1, median at 20, Q3 at 25, maximum at 50. The bottom box
Identify any values of data that might affect the statistical measures of spread and center. (2 points)

The females worked less than the males, and the female median is close to Q1.
There is a high data value that causes the data set to be asymmetrical for the males.
There are significant outliers at the high ends of both the males and the females.
Both graphs have the required quartiles.
7. (06.02)
The table below shows data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:

High Low Q1 Q3 IQR Median Mean σ
College 50 6 8.5 17 8.5 12 15.4 11.7
High School 28 3 4.5 15 10.5 11 10.5 5.8

Which of the choices below best describes how to measure the spread of this data? (2 points)

Both spreads are best described with the IQR.
Both spreads are best described with the standard deviation.
The college spread is best described by the IQR. The high school spread is best described by the standard deviation.
The college spread is best described by the standard deviation. The high school spread is best described by the IQR.

Answers

Final answer:

1. The dataset with the smallest IQR is the Mid-term exams. 2. The exam median is only 1–2 points higher than the class median. 3. The standard deviation is the best measurement of spread for games and movies.

Explanation:

1. The dataset with the smallest IQR is the Mid-term exams. To find the IQR, first, calculate the first quartile (Q1) and the third quartile (Q3). Then, find the difference between Q3 and Q1. By comparing the IQR values for the Mid-term and Final exams, it can be determined that the Mid-term exams have the smallest IQR.

2. The exam median is only 1–2 points higher than the class median. The box plot's median represents the middle value of the dataset. By comparing the medians of the exam and class data, it can be determined that the exam median is only 1–2 points higher than the class median.

3. The standard deviation is the best measurement of spread for games and movies. While the IQR can measure spread, the standard deviation is a more precise measurement. Comparing the spread of the data, the standard deviation is the best measurement for both games and movies.

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Final answer:

Various statistical measures such as IQR, Standard Deviation, Median and Mean were used to interpret the given data in numerous scenarios. Outliers were identified and impacts on datasets were evaluated.

Explanation:

To answer these questions, we need to understand few key statistical terms: 'Mean' is the simple average of data, 'Median' is the middle score of data, 'IQR' (Inter Quartile Range) is the difference between the upper quartile (Q3) and the lower quartile (Q1), which helps in understanding the spread and '.'Standard Deviation' measures the absolute variability of a dataset.

Question 1: IQR for the mid-term exams is Q3 (92) - Q1 (82) = 10. IQR for the final exams is Q3 (85) - Q1 (78) = 7. So, the final exams have the smallest IQR.

Question 2: The boxes showing median indicates that the exam median is only 1–2 points higher than the class median.

Question 3: Since the spread of the data at basketball games and local movie theaters demonstrates varied distributions, both the IQR and the standard deviation should be used to evaluate the data spread.

Question 4: Box plots do not provide direct information on the mean. So, there is no way of telling what the means are.

Question 5: For the video hours, we see that values 18, 21, and 26 lie far from the main part of the data, thus they can be considered as possible outliers.

Question 6: The male data set shows a high data value, which causes the data set to be asymmetrical. This could affect statistical measures like the mean and standard deviation.

Question 7: The spread of the college data, having a large standard deviation and IQR, is best described by the standard deviation. The high school data, with smaller numbers, is best described by the IQR.

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Can someone help me solve 2√x−4=10 I'm confused.

Answers

let's analyze two cases

case A) 2(√x)−4=10
2√x=10+4----> 2√x=14----> √x=14/2----> square---> x=7²----> x=49

case B)
2√(x-4)=10
√(x-4)=10/2-----> square---> x-4=5²----> x=25+4-----> x=29

The cost of parking in a garage, in dollars, can be modeled by a step function whose graph is shown. How much does it cost to park for 3 hours and 45 minutes?

$3
$4
$6
$10

Answers

3 hours 45 minutes=3 hours (45 minutes)(1 hour)/(60 minutes)
3 hours 45 minutes=3 hours 0.75 hours
3 hours 45 minutes = 3.75 hours
3 hours < 3.75 hours < 4 hours →Cost=$6

Answer: Third option $6

Answer:

the cost is [tex]\$6[/tex]

Step-by-step explanation:

observing the graph

we know that

For the interval of x------> [tex](0,1][/tex] -----> the cost is [tex]\$0[/tex]

For the interval of x------> [tex](1,3][/tex] -----> the cost is [tex]\$4[/tex]

For the interval of x------> [tex](3,4][/tex] -----> the cost is [tex]\$6[/tex]

For the interval of x------> [tex](4,infinite)[/tex] -----> the cost is [tex]\$10[/tex]

In this problem we have

[tex]3[/tex] hours and [tex]45[/tex] minutes

therefore

the value of x belong to the interval [tex](3,4][/tex]

the cost is [tex]\$6[/tex]

a quarterback for the seattle seahawks completes 54% of his passes. let the random variablle x be the number of passes completed in 20 attempts. conduct a stimulation of the 20 attempts using the following random digits. be sure to state how your assign your didgits

Answers

The complete question is:
"A quarterback for the Seattle Seahawks completes 54% of his passes. Let random variable X be the number of passes completed in 20 attempts. Conduct a simulation of the 200 attempts using the following random digits. Be sure to state how you assign your digits. 98726 10983 56239 42042 76520 68276 58239 48729 84912 87491

a) What proportions of passes were completed?

b) How does this compare with what “should have happened” theoretically?"

First, take 100 numbers, i.e. from 00 to 99, and set that the first 54% are completed passes, therefore from 00 to 53, and the remaining 46% are not completed passes, therefore from 54 to 99.

Now, divide the random digits until you compose 20 numbers: 98, 72, 61, 09, 83, 56, 23, 94, 20, 42, 76, 52, 06, 82, 76, 58, 23, 94, 87, 29.

Then, divide the numbers into the two set categories:

completed: 09, 23, 20, 42, 52, 06, 23, 94, 87 = 9 outcomes
not completed: 98, 72, 61, 83, 56, 94, 76, 82, 76, 58, 29 = 11 outcomes

A) Therefore, from the random digits you get 9/20 = 45% of completed passes and 11/20 = 55% of not completed passes.

B) Using the random digits, the simulation gave an outcome exactly opposite to what we expected.

What is the gcf of 3x^2 and 7y

Answers

Answer :
Unless we have a relationship defined between x and y there is no common factor apart from 1.

hope it helps
~Reema

Which expressions are equivalent to the one below? Check all that apply.

16^x/4^x

A. 16^x

B. (16-4)^x

C. 4^x * 4^x/ 4^x

D. (16/4)^x

E. 4^x

F. 4

Answers

16^x / 4^x
= 4^2x / 4^x
= 4^x so E is one choice.
also C and D

Math help please!!!!! If AO = 21 and BC = 14, what is AB?

Answers

the answer is 28 because line BO is similar to AB

Given is a circle O with tangent AB and secant OB.

Given is OA = 21 units and BC = 14 units.

From the diagram, OB = OC + BC.

OC and OA, both are radius, so OC = OA = 21 units.

Now OB = 21 + 14 = 35 units.

In right triangle ΔOAB, using Pythagorean theorem;

OA² + AB² = OB²

⇒ (21)² + AB² = (35)²

⇒ 441 + AB² = 1225

⇒ AB² = 1225 - 441 = 784 square units

⇒ AB = [tex]\sqrt{784} =28[/tex]

⇒ AB = 28 units.

Hence, final answer is AB = 28 units.

Find the mean, median, and mode
15, 3, 11, 15, 1, 14, 7, 2, 1, 1, 2

A. mean = 6.5, median = 8, mode =1
B. mean = 6, median = 3, mode = 1
C. mean = 6, median = 3, mode = 8
D. mean = 6.5, median = 3, mode = 1

Answers

D is the answer I got

mean=6.5
median=3
mode=1
so the answer is d
I know its right for sure :)

Please help! Math question

Answers

Your arithmetic sequence has first term 2 and common difference 3. The rule is
[tex]a_{n}=a_{1}+d(n-1)\\a_{n}=2+3(n-1)\\a_{n}=3n-1[/tex]

The graph shows the first 8 terms.

Factor 2x2 - 11x - 21.
A) (2x + 3)(x - 7)
Eliminate
B) (x + 3)(2x - 7)
C) (2x - 3)(x + 7)
D) (2x + 7)(x - 3)

Answers

The first step for factoring this expression is to write -11x as a difference.
2x² + 3x - 14x - 21
Factor out x from 2x² + 3x.
x × (2x + 3) - 14x - 21
Now factor out -7 from -7(2x + 3)
x × (2x + 3) - 7(2x + 3)
Finally,, factor out 2x + 3 from the expression.
(2x + 3) × (x - 7)
This means the correct answer to your question is (2x + 3) × (x - 7),, or option A.
Let me know if you have any further questions.
:)

Need help with these 2 questions please and thanks

Answers

These are two questions and two answers.

1) Problem # 15.

(i) [tex] \lim_{x \to \ 60^{-} } f(x) [/tex]

You have to approach the value of x by the left. That is the horizontal segment at y = 56 (with the open circle to the left and the solid circle to the right).

So, the value of the limit es 56.

(ii) [tex] \lim_{x \to \ 60^{+} } f(x)[/tex]

You have to approach the x value by the right. That is the horizontal segment at y = 68.

So the value of the limit is 68.

(iii) Since, the existence of the limit requires that both limits from the left and the right be equal, the conclusion is that the limit does not exist.

That is shown in the graph because the way the function is defined is different if the value of x is greater or less than 60.

(iv) What causes the graph to jump vertically by the same amount at discontinuites is that the rates are defined in intervals and not in a continuous way.

2) Problem 16. Mathematical induction.

Mathematical induction requires 3 steps: 1) Initial hypothesis, 2) assume the equation is valid for n = k, and 3) prove the equation is valid for n = k+1.

This is the solution step by step.

1) Initial hypothesis: first term ⇒ n = 1

left side: 8
rigth side: 4n(n+1) = 4(1+1) = 4(2) = 8

8 = 8 ⇒ check

2) Assume n = k

letf side: 8 + 16 + 24 + ... + 8k
right side: 4k(k + 1)

3) Proove for n = k + 1

left side: 8 + 16 + 24 + ... + 8k + 8(k+1)
right side: 4(k+1) (k + 1 + 1) = 4(k+1)(k+2)

Since 8 + 16 + 24 + ... 8k is assumed to be equal to 4k(k+1), then the left side ends as:

left side: 4k(k+1) + 8(k+1)

take common factor k + 1 → (k+1)(4k + 8)
take common factor 4 for the second parenthesis → (k+1) 4(k + 2) = 4(k+1)(k+2)

So, we have shown that the left side is equal fo the right side, which completes the proof by induction.

Find the fifth term of the arithmetic sequence in which
t1 = 3 and tn = tn-1 + 4.

A) 5
B) 7
C) 19
D) 23

Answers

We have given [tex] t_{1}=3 [/tex] and [tex] t_{n} = t_{n-1} + 4 [/tex]. Fifth term is [tex] t_{5}= t_{4}+4 [/tex]. It means we need to find [tex] t_{4} [/tex] and in order to find the latter, we have to find [tex] t_{4}= t_{3}+4 [/tex], [tex] t_{3}= t_{2}+4 [/tex] and [tex] t_{2}= t_{1}+4 [/tex] . Since we know the value of [tex] t_{1} [/tex], beginning from the last equation we can find the value of [tex] t_{5} [/tex]. We can write that [tex]t_{2}= t_{1}+4=3+4=7[/tex], [tex]t_{3}= t_{2}+4=7+4=11[/tex] and [tex]t_{4}= t_{3}+4=11+4=15[/tex]. Since [tex] t_{5}= t_{4}+4 [/tex], then [tex] t_{5}= 15+4=19 [/tex]

Mrs. Jones Algebra 2 class scored very well on yesterday's quiz. With one exception, everyone received an A. Within how many standard deviations from the mean do all the quiz grades fall?

91, 92, 94, 88, 96, 99, 91, 93, 94, 97, 95, 97

A. 1
B. 2
C. 4
D. 3

Answers

Answer: option B. 2

Explanation:

1) Find the mean:

mean = sum of the values / number of data

The number of data is 12, so:

mean = (91 + 92 + 94 + 88 + 96 + 99 + 91 + 93 + 94 + 97 + 95 + 97) / 12

mean = 1127 /12 = 93.917

2) Find the variance:

variance = sum of squares of the differences between each data and the media, divided by thenumber of data - 1

variance = (106.97)/(12-1) = 106.97 / 11 = 9.720

3) Find the standard deviation

standard deviation = square root of variance = √(9.720) = 3.118

4) Find the difference of the maximum and the minimun grades with the media:

maximum grade: 99 - 93.917 = 5.083

Find how many standard deviations that is: 5.083 / 3.118 = 1.63

minimum grade: |88 - 93.917| = 5.917

Find how many standard deviations that is: 5.917 / 3.118 = 1.9

5) Conclusion: all the quiz grades falls inside 2 standard deviations form the mean.

Factor 1/2 out of 1/2z+9

Answers

1/2(z+18) is the answer i believe

1/2(z+18) is the expression of 1/2z+9 when 1/2 is factored out.

What is Fraction?

A fraction represents a part of a whole.

Given,

The expression is 1/2 z+9

A factor is a number that divides another number, leaving no remainder.

The given expression is one by two times of z plus nine.

1/2 z+9

Now we need to take 1/2 as common from the expression

1/2(z+18)

Hence, 1/2(z+18) is the expression of 1/2z+9 when 1/2 is factored out.

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The length of a rectangle is four times its width. if the width is 15, what is the area?

Answers

15*4= 60
You multiply the width by four to get the answer.
60*15 = 900
The area is 900.

Use formulas to find the lateral area and surface area of the given prism. Round your answer to the nearest whole number.

Answers

Lateral area = Perimeter x height

Lateral area = (2 + 7 + 7.28) x 31 = 16.28 x 31 = 505 m²

Surface area = 2(1/2 x 2 x 7) + 505 = 519 m²

Answer: Lateral area = 505 m² ; Surface area = 519 m²

Someone who knows math more than I do, can you please answer this question for me, I'd appreciate it so much <3

A right triangle has one angle that measures 28o. The adjacent leg measures 32.6 cm and the hypotenuse measures 35 cm.

What is the approximate area of the triangle? Round to the nearest tenth.

Area of a triangle = 1/2 bh

Answers

Using the Pythagorean theorem, the other leg measures:
x = sqrt(35^2 - 32.6^2) = 12.74 cm
Since this is a right triangle, the base and height are simply the two sides that are perpendicular to each other.
Then the approximate area is (1/2)(leg 1)(leg 2) = (1/2)(32.6)(12.74) = 207.6 cm^2.

Factor this expression.

3x2 – 6x

A. 3(x – 2)
B. 3(x2 – 2)
C. 3x(x2 – 2)
D. 3x(x – 2)

Answers

The factor will result to answer which is D

Answer:

D.3x(x-2)

Step-by-step explanation:

i already did this

Karl and his dad are building a playhouse for karl's younger sister. the floor of the playhouse will be a rectangle that is 6 by 8 1/2 feet. how much carpeting do karl and his dad need to cover the floor.

Answers

Area of a rectangle:
[tex]l \times w[/tex]
Area of playhouse"
[tex]6 \times 8 \frac{1}{2} \\ = 51 {ft}^{2} [/tex]

The playhouse floor is a rectangle with dimensions of 6 by 8.5 feet. To determine the amount of carpet needed, multiply the length and width to calculate the area, which is 51 square feet.

Karl and his dad need to know the amount of carpeting required to cover the floor of a playhouse, which is a practical mathematics problem involving area calculation. To find out how much carpet they need, they have to calculate the area of the rectangular floor, which is the product of its length and width.

The floor measures 6 feet in length and 8.5 feet in width. Multiplying these two dimensions gives us the area:

Area = Length times Width

Area = 6 ft times 8.5 ft

Area = 51 square feet

Therefore, Karl and his dad would need to purchase 51 square feet of carpeting to cover the playhouse floor.

What is the correct evaluation of 15-x, when x is equal to -5?

Answers

15-(-5)=20 because the negative and minus sign cancel each other out

2. The height of one square pyramid is 24 m. A similar pyramid has a height of 8 m. The volume of the larger pyramid is 648 m3. The surface area of the smaller pyramid is 124 m^2. Determine each of the following, showing all your work and reasoning: find the surface area of the larger pyramid

Answers

Volume of square pyramid = a²h / 3

Surface area of square pyramid = a² +2a √((a²/4) + h²)

a = base edge
h = height

for the large pyramid;
volume = 648 m³
height = 24 m

Volume = a²h / 3
648 m³ = a² x 24 m / 3
a² = 81 m²
a = 9 m

by applying surface area equation to the large pyramid,
surface area = a² +2a √((a²/4) + h²)
= 9² + 2 x 9 √((9² / 4 ) + 24²
= 520.53 m²

hence surface area of large pyramid = 520.53 m²

V = a²h / 3

V = a²h / 3648 * 3 / 24= a² a² = 81 a = 9
Area of square is 9 sq.m.

Area of the lateral sides of the larger pyramid:
a = 1/2 b*h
b = 9
h = sqrt (24^2 + 4.5^2)
h = 24.42
a = 1/2 (9 * 24.42) = 109.88

SURFACE AREA = 4 lateral sides area + base area
SURFACE AREA = 4 * 109.88 + 9*9 = 520.53

Every year, 50 million flea collars are thrown away. How many flea collars are thrown away per day, rounded to the nearest thousand

Answers

Flea collars thrown away per year = 50 million

Flea collars thrown away per day = 50 million/Average number of days in a year

Average number of days in a year = 365 days
Therefore,
Flea thrown away per day = 50,000,000/365 = 136,986.3

Rounding the answer to nearest thousand,
Number of flea thrown away per day = 137,000

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