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3 years ago

7

A cereal box is a rectangular prism that is 8 inches long and 2 1/2 inches wide. the volume of the box is 200 inches cubed. what

is the height of the box?

1 answer:

Archy [21]3 years ago

8 0

You have to multiply 8 and 2 1/2 which is 20. Then divide 20 and 200. Then you get 10 and that is your answer.

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The height of a cylinder is 5 centimeters. The circumference of the base of the cylinder is 16 centimeters. Find the lateral sur

ad-work [718]

The lateral surface area of the cylinder = 80 sq cm.

Step-by-step explanation:

Given,

The circumference of the base of the cylinder is 16 cm

So, 2 X pi X r = 16 cm ( where r be the radius of the base)

Now, the lateral surface of the cylinder = 2 X pi X r X H ( where H be the height = 5 cm)

= 16 X 5 sq cm

= 80 sq cm

Hope it helps you.

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3 years ago

How much will 2 3/4 yards of fabric cost at $2.99 per yard

kipiarov [429]

2 3/4 * 2.99 =
11/4 * 2.99 =
32.89/4 =
8.22 <==

or
2 3/4 = 2.75
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3 years ago

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If f(x)=6x^2-4 what is f(-1)

Nana76 [90]

Answer:

2

Step-by-step explanation:

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3 years ago

How many dimes did she have

Marina CMI [18]

She had 15 dimes. That's the answer

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3 years ago

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Natasha2012 [34]

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$ are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

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

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

5 0

3 years ago

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