What is the area of the figure below ?
A. 12
B. 15
C. 18
D. None of these

Answer:

B A C A B

Step-by-step explanation:

i took this test before

Answer:

mean

Step-by-step explanation:

The mean will decrease, in an option of mean, mode, median and range:)

Using the given linear function, it is found that Avery was 7 kilometers above sea level before she started climbing.

A linear function is modeled by:

[tex]y = mx + b[/tex]

In which:

In this problem, her altitude in kilometers after t hours is modeled by:

A(t) = 7 + 1.3t.

The y-intercept is of b = 7, which means that Avery was 7 kilometers above sea level before she started climbing.

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The car on a Ferris wheel with a radius of 20 feet. travels a distance of 125.6 feet.

In order to determine the distance that the car has travelled, we will use the formula for calculating the circumference of a circle.

C = 2πr

where

r = radius

Therefore,

r is the radius of the fairy wheel

r = 20 ft

C = 2 × 3.14 × 20

C = 40 × 3.14

C = 125.6 ft

This shows that the car on a Ferris wheel with a radius of 20 feet. travels a distance of 125.6 feet

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Answer:

No Solution

Step-by-step explanation:

-y = -2x + 8

y = 2x - 8

2x + 8 = 2x - 8

2x - 2x = -8 - 8

0 = -16

No Solution

You can also graph y = 2x + 8 and y = 2x - 8. The slope are the same and so they are parallel

Answer:

⠀

Step-by-step explanation:

⠀

We know,

[tex]{\longrightarrow \qquad \boldsymbol{\pmb{Area_{(semicircle)} = \dfrac{1}{2}( \pi {r}^{2} ) }}}[/tex]

⠀

Where,

⠀

Now, Substituting the values in the formula :

⠀

[tex]{\longrightarrow \qquad \rm{{Area_{(semicircle)} = \it \dfrac{1}{2} \times 3.14 \times ({8})^{2} }}}[/tex]

[tex]{\longrightarrow \qquad \rm{{Area_{(semicircle)} = \it \dfrac{1}{ \cancel2} \times 3.14 \times \cancel{64} }}}[/tex]

⠀

[tex]{\longrightarrow \qquad \rm{{Area_{(semicircle)} = \it {1} \times 3.14 \times 32 }}}[/tex]

⠀

[tex]{\longrightarrow \qquad \rm{{Area_{(semicircle)} = \it {1} \times 100.48 }}}[/tex]

⠀

[tex]{\longrightarrow \qquad \boldsymbol{ \pmb{Area_{(semicircle)} \approx \it 100.48 }}}[/tex]

⠀

Therefore,

Answer:

100.48 cm² ( option 2 )

Step-by-step explanation:

Given:-

To find:-

Solution:-

Area of semi-circle :-

[tex] \frac{1}{2} \pi \: r {}^{2} [/tex]

1/2 × 3.14 × 8²

[tex]100.48 \: cm {}^{2} [/tex]

Answer:

point A is the answers for the question

Step-by-step explanation:

please mark me as brainlest

Answer:

A) 9 and 15.

Step-by-step explanation:

First, let's see what each number's factor is:

A)

9 : 1, 3 & 9

15 : 1, 3, 5 & 15

B)

6 : 1, 2, 3 & 6

10 : 1, 2, 5 & 10

C)

8 : 1, 2, 4 & 8

12 : 1, 2, 3, 4, 6 & 12

Therefore, The number 3 is a common factor of A) 9 and 15.

___

Answer:

a = 7

Step-by-step explanation:

3x + 4 = 57 ÷ 3

3x + 4 = 19

3x = 15

x = 5

2x + 2 =  2(5) + 2 = 10 + 2 = 12

19 - 12 = 7

a = 7

hope this helps!! p.s. i really need brainliest :)

Answer:

Step-by-step explanation:

The largest number of groups they can be split into and have the same number of boys and girls on each team is 6 groups

Given:

Number of girls = 12

Number of boys = 16

Find the highest common factor of 12 and 16

12 = 1, 2, 4

18 = 1, 2,4

The highest common factor of 12 and 16 is 4

Number of girls in each group = 12/4

= 3 girls

Number of boy in each group = 18/4

= 4.5 boys=5

Therefore, the group can be split into 4 groups to have 3 girls and 5 boys in each group

Answer:

the quotient of 18 and 3 is 6

Step-by-step explanation:

do the inverse and get 6x3=18

Answer:

True I'm pretty sure. Just check to be sure.

Step-by-step explanation:

The length of the wire is the perimeter of the wire

The radius of the circle is 1.59 meters

The length of the wire is given as:

L  = 20m

When cut into two pieces, we have:

The circumference of the circle is calculated using:

C = 2πr

So, we have:

2πr = 10

Make r the subject

r = 10/2π

Evaluate the quotient

r = 1.59

Hence, the radius of the circle is 1.59 meters

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The Pythagorean theorem is the idea that the sum of the two legs which are both squared is equal to the hypotenuse's length squared.

     *look at the image I attached for a better explanation

By looking at the picture, we are missing the leg's length, and by using the Pythagorean theorem, we get the equation

    [tex]?^2+5^2 = 13^2\\?^2 + 25 = 169\\?^2 = 144\\? = 12[/tex]

Thus the missing side length is 12 cm

Hope that helps!

Answer: 3005

Step-by-step explanation: multiply 5 by 20 and it will give 100 then times 30 Equals 3000 and plus gives 3005 . Please mark brainliest

Answer:3005

Step-by-step explanation :multiply 5 by 20 and it will give 100 then multiply by 30  that Equals 3000 and plus 5 that gives you 3005

Answer:

The first x=4       y=8

Step-by-step explanation:

2x-5 = y-5

x+1 = y-3

  Multiply the second by -2(to eliminate x)

  2x-5 =y-5

 -2x-2= -2y+6

      ADD BOTH

-7 = -y+1

y=8

            PLUG y

x+1 =y-3

x+1 =8-3

x+1=5

x=4

Answer:

Label ELabel E

Step-by-step explanation:

Label ELabel ELabel k

Answer:

T. Perimeter

Step-by-step explanation:

The miles Peyton's family had travelled when he fell asleep is 770 miles

Percentage can be described as a fraction out of an amount that is usually expressed as a number out of hundred. The sign used to represent percentages is %.

Miles when Peyton fell asleep = 77% x 1000 = 770 miles

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Answer:

Step-by-step explanation:

Holding the perimeter fixed, the shape with greatest area is square ( all sides of the same) length) For your name example a square with a perimeter of 20 cm has sides of 5 cm and area of 5 times 4=20 cm) ^3 hope this helps

Answer:

9n/60

Step-by-step explanation:

add 1/12 + 1/15 which = 5/60 + 4/60.

= 9/60

YAY!!!!!!!

The fraction of the exams that Mr Tough and Mrs Nice grade in 1 hour if they work together is 3/20

In Maths, a fraction is used to represent the portion/part of the whole thing. A fraction has two parts, namely numerator and denominator. There are proper fractions, improper fractions and mixed fractions.

Example: 2/5, 1/4 etc.

We will this equation,

rate × time = work done

For this problem:

             Mr Tough's rate:

                          Mr Tough's rate × 12 hours = final exam

                          Mr Tough's rate = 1/12

             Mrs.Nice's' rate:

                            Mrs .Nice's' rate × 15 hours = final exam

                            Mrs .Nice's' rate = 1/15

To make this into a solvable equation,  find the total time (T) needed for Mr Tough and Mrs.Nice to grade the final exam. This time is the sum of the rates of Mr Tough and Mrs.Nicea, or:

Total time:

T(1/12 hours + 1/15 hours) = final exam

T =  final exam / (1/12 + 1/15)

Simplifying we get,

T =  final exam / (0.15)

Now, to finish in 1 hour we will find what fraction of the final exam they can grade, so taking T = 1 hour,

1 hour =  final exam / (0.15)

By further simplifying,

final exam = 1 × 0.15

                 = 0.15

                 = 15/ 100

                 = 3/20

Therefore, 3/20 fraction of the final exam they can grade if they work together for 1 hour.

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Answer:

20 and 13

Step-by-step explanation:

If one list is 'a' and the other one is 'b' then

a-b = 7 and a+b = 33

a-b+a+b = 7+33

2a = 40

a = 20

Now, 20 - b = 7

-b = -13

b = 13

Answer:

Question 3

Given expression:

[tex]\sf \left(\dfrac{-8}{9} \times \dfrac{1}{-5}\right)+\left(\dfrac{-8}{9} \times \dfrac{-7}{11}\right)[/tex]

Both parentheses share a common term.

To factor out a common term: ab + ac = a(b + c)

[tex]\textsf{Factor out the common term }\dfrac{-8}{9}[/tex]

[tex]\sf \implies\dfrac{-8}{9} \times \left(\dfrac{1}{-5}+\dfrac{-7}{11}\right)[/tex]

-------------------------------------------------------------------------------------------------

Question 4

To calculate the amount of money Rajni had to begin with, sum the three amounts she spent:

[tex]\sf \implies total=10\frac14+25\frac34+16\frac12[/tex]

As [tex]\sf a\frac{b}{c}=a+\frac{b}{c}[/tex], then:

[tex]\sf \implies 10+\frac14+25+\frac34+16+\frac12[/tex]

Collect and sum the whole numbers:

[tex]\sf \implies (10+25+16)+\frac14+\frac34+\frac12[/tex]

[tex]\sf \implies 51+\frac14+\frac34+\frac12[/tex]

As  [tex]\sf \frac14+\frac34=\frac44=1[/tex], then

[tex]\sf \implies 51+1+\frac12[/tex]

[tex]\sf \implies 52+\frac12[/tex]

[tex]\sf \implies 52\frac12[/tex]

The cost to buy the flooring for Raul's bedroom is $364.65.

A scale drawing is a reduced form in terms of dimensions of an original image / building / object

Actual length of the bedroom = 4 x 3.25 = 13 feet

Actual length of the bedroom = 4 x 2.75 = 11 feet

Area = 143 square feet

Cost = 143 x 2.55 = $364.65

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Answer:

86.5

Step-by-step explanation:

Length of the string forms the hypotenuse AC of right angle triangle ABC.

C is the kite. A is the hand of the person (or, one end of the string). CB is the perpendicular

(height of the kite) from C on the ground. BC is the horizontal.

Sin 60° CB/AC = √3/2

height = 1.73/2 AC = 0.865 * 100 m = 86.5 m

Answer:

㏒₆ (18) - ㏒₆ (6) = ㏒₆ (3)Here subtraction of logarithm given. So we have to use the property of logarithm of quotient.The property says that㏒ₐ (M) - ㏒ₐ (N) = ㏒ₐ (M/N)So for subtraction we have to divide them. That is logarithm of difference becomes the quotient.So we can write,㏒₆ (18) - ㏒₆ (6) = ㏒₆ (18/6) If we divide 18 by 6 we will get 3.So, ㏒₆ (18) - ㏒₆ (6) = ㏒₆ (3)

Step-by-step explanation:

Answer:Quotient Property

Step-by-step explanation:The property used to rewrite the expression is the quotient rule of logarithms, which states that log_b(x/y) = log_b(x) - log_b(y).

The volume of the chocolate is the amount of choclate used in the cherries

The amount of chocolate used is 17.24 cubic centimeters

Start by calculating the volume of the cherries using:

V = 4/3 π(D/2)^3

The diameter of the cherry is 2.8 cm.

So, we have:

V = 4/3π(2.8/2)^3

Evaluate

V = 11.49

Next, calculate the volume of the cherry and the chocolate.

V = 4/3π(0.5+2.8/2)^3

V = 28.73

Calculate the difference (d) between both volumes

d = 28.73 - 11.49

d = 17.24

Hence, the amount of chocolate used is 17.24 cubic centimeters

Read more about volumes at:

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Step-by-step explanation:

it clearly says (and also shows it by specifying 4cm × 4cm side lengths) : a square base.

so, all net diagrams not having a square in the middle (but a rectangle or a triangle) are out.

and in fact, all 3 that have a square in the middle, are valid nets that would fold up to the pyramid : the 2nd, 4th and 6th.