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White raven [17]

3 years ago

15

The volume of a cone is 31.4 cm3 and the radius is 2 cm. What is the height of the cone? Use 3.14 for π. Round to the nearest te

nth, if needed.
A) 22.5 cm
B) 7.5 cm
C) 15 cm
D) 41.9 cm

2 answers:

kupik [55]3 years ago

8 0

It's (B) 7.5 cm

h= (3V) / (πr^2)

dedylja [7]3 years ago

8 0

Hello There!

The formula is 3 (v/πr²)
Substitute the values in:
3 (31.4/π2²)
Solve:
The height is 7.5cm.
Therefore, making your answer B.

Hope This Helps You!
Good Luck :)

- Hannah ❤

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Equation 1: y = -2x + 1

Equation 2: y = 2x - 3

Since both equations already have y isolated, we are able to simply set the right side of both equations equal to each other. Since we know that the value of y must be the same, we can do this.

-2x + 1 = 2x - 3

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Then, we need to plug our value of x back into either of the original two equations and solve for y. I will be plugging x back into equation 2 above.

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3 years ago

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Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

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3 years ago

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What is the solution of the system of equations? {7/2x−1/2y=9/23x−y=5 Enter your answer in the boxes. ( _ , _ )

dimulka [17.4K]

Answer:

(1, -2)

Step-by-step explanation:

I hope that helps :/

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Answer:

7:17 PM

Step-by-step explanation:

Add another 30 minutes to 6:30 = 7:00 PM

47-30 = 17

add the remaining 17 minutes to 7:00 = 7:17 PM

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