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3 years ago

I really neeed help!! 20 points!!,

Mathematics

1 answer:

Lesechka [4]3 years ago

5 0

Try this solution:
V=14*11*7+5*5*7=1078+175=1253 in³

answer: 1253 in³

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Rob ate 3 cookies and drank 2 cups of milk. Barry ate 8 cookies and drank 1 cup of milk. Kevin ate 9 cookies and drank 4 cups of

harkovskaia [24]

<h3><u>Cookies</u></h3>

Rob: 3

Barry: 8

Kevin: 9

Let's add all of that up.

3 + 8 + 9

11 + 9

= 20 cookies total

Rob: 2

Barry: 1

Kevin: 4

Let's add this up.

2 + 1 + 4

3 + 4

= 7 cups of milk

They had 20 cookies and 7 cups of milk before they ate.

8 0

4 years ago

Read 2 more answers

An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or abov

babunello [35]

Answer:

D) 14 seconds

Step-by-step explanation:

First we will plug 500 in for y:

500 = -4.9t² + 120t

We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:

500-500 = -4.9t² + 120t - 500

0 = -4.9t²+120t-500

Our values for a, b and c are:

a = -4.9; b = 120; c = -500

We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.

The quadratic formula is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using our values for a, b and c,

$x=\frac{-120\pm \sqrt{120^2-4(-4.9)(-500)}}{2(-4.9)}\\\\=\frac{-120\pm \sqrt{14400-9800}}{-9.8}\\\\=\frac{-120\pm \sqrt{4600}}{-9.8}\\\\=\frac{-120\pm 67.8233}{-9.8}\\\\=\frac{-120+67.8233}{-9.8}\text{ or }\frac{-120-67.8233}{-9.8}\\\\=\frac{-57.1767}{-9.8}\text{ or }\frac{-187.8233}{-9.8}\\\\=5.3242\text{ or }19.1656$

The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is

19-5 = 14 seconds.

6 0

3 years ago

A 10ft by 20ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216ft^

Dmitry [639]

check the picture below.

so, we know the dimensions of the pool, is a 20x10, so its area is simply 200 ft², and we know the walkway is 216 ft², so the whole thing, including pool and walkway is really 200 + 216 ft².

now, as you see in the picture, the dimensions for the combined area is 20+2x and 10+2x, since the walkway is "x" long, therefore,

$\bf \stackrel{length}{(20+2x)}\stackrel{width}{(10+2x)}=200+216\implies \stackrel{FOIL}{4x^2+60x+200}=200+216 \\\\\\ 4x^2+60x=216\implies \stackrel{dividing~by~4}{x^2+15x=54} \\\\\\ x^2+15x-54=0\implies (x+18)(x-3)=0\implies x= \begin{cases} -18\\ \boxed{3} \end{cases}$