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3 years ago

What is the answer ?

Mathematics

1 answer:

WINSTONCH [101]3 years ago

3 0

Ok, hopefully this is correct but bill is faster so Kevin is slower by ten because bill was going 110 at 2 hours which divided is 55 and Kevin was going 90 at 2 hours which is 45. Hope this helped! Have a great day :D

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The frequency table will be used to make a histogram. Use the drop-down menus to answer each question regarding the histogram.

Brut [27]

Answer:

The frequency table will be used to make a histogram. Use the drop-down menus to answer each question regarding the histogram.

What will the horizontal axis represent?

✔ temperature

Which interval should be used for the horizontal axis?

✔ 20

What will the vertical axis represent?

✔ number of days

What is an appropriate scale for the vertical axis?

✔ 0–100

Step-by-step explanation:

Hope this helps

6 0

3 years ago

2x ( 4x - 5) (22 – 81) = 0

shepuryov [24]

Answer:

0.500

Step-by-step explanation:

6 0

3 years ago

Aysha’s score for a game is given by the expression 9 − 8.2 − (−1.9).

11Alexandr11 [23.1K]

Answer:

Aysha

Step-by-step explanation:

Aysha's score = 9 – 8.2 = 0.8 + 1.9 = 2.7

Beth's score = –8 + 5.4 = (–2.6) – 1.8 = – 4.4

2.7 > (– 4.4)

3 0

2 years ago

This question has three parts. Answer the parts in order.

nalin [4]

Answer:

The area of the smallest section is $A_{1}=100yd^{2}$

The area of the largest section is $A_{2}=625yd^{2}$

The area of the remaining section is $A_{3}=250yd^{2}$

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

$A_{1}=10yd*10yd$

$A_{1}=100yd^{2}$

- Area of the second section:

$A_{2}=x^{2}$ (Eq.1)

- Area of the third section:

$A_{3}=width*length$

$A_{3}=10yd*x$ (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

$A_{1}+A_{2}+A_{3}=975yd^{2}$

Replacing values:

$100+x^{2}+10x=975$

Solving for x:

$x^{2}+10x-875=0$

$x=\frac{-10+\sqrt{100+(4*875)}}{2}$

$x=\frac{-10+\sqrt{3600}}{2}$

$x=\frac{-10+60}{2}$

$x=25$

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

$A_{2}=25^{2}$

$A_{2}=625yd^{2}$

- From Eq.2:

$A_{3}=10*25$

$A_{3}=250yd^{2}$

3 0

3 years ago

Help me please, kinda lost

Dmitry_Shevchenko [17]

Answer:

400cm^2

Step-by-step explanation:

For this question, you need to find the surface areas of both objects, then add them together.

For the cube:

56cm for the front

56cm for the back

42cm for the side

42cm for the other side

48cm for the top

48cm for the bottom

SA for the cube: 292cm^2

For the triangle, things are a little different. We do the same process that we did to find the SA of the cube, but we do not have to find the SA of the top side (because there isn't a top side) and we have to divide the total SA by 2 because it's a triangle.

For the triangle:

42cm for the front

42cm for the back

54cm for the slanted side

42cm for the other side

36cm for the bottom

SA for the triangle: (216/2)^2=108cm^2

292+108=400cm^2

the SA of the composite figure is 400cm^2

8 0

3 years ago