algebra problems

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Elaine

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Sep 10, 2005
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identify the vertex of the parabola

f(x)=-7^2 ?

Is it (-7,0).....


#2

the intensity of I light varies inversely as the sqiare of the distance D from the source. If the intensirty of illumination on a screen 5ft. from a light is 2 foot candles , what's the intensity on a screen 15 foot from the light?

Is the soultion 2/10 foot candle??

thankyou!!!
 
identify the vertex of the parabola

f(x)=-7^2 ?
Click to expand...

Is this supposed to be f(x)=-7x^2 ?

Ask yourself, "How does the coefficient of x affect the graph?" Try plotting a couple of graphs with different coefficients and see the effect.
 
identify the vertex of the parabola

f(x)=-7^2 ?
Click to expand...

Is this supposed to be f(x)=-7x^2 ?

Ask yourself, "How does the coefficient of x affect the graph?" Try plotting a couple of graphs with different coefficients and see the effect.
Click to expand...

Yes, it is supposed to be that.
 
identify the vertex of the parabola

f(x)=-7^2 ?

Is it (-7,0).....

wrong

it's 0,0......


#2

the intensity of I light varies inversely as the sqiare of the distance D from the source. If the intensirty of illumination on a screen 5ft. from a light is 2 foot candles , what's the intensity on a screen 15 foot from the light?

Is the soultion 2/10 foot candle??

okay it's 2.9 got it...

thankyou!!!
Click to expand...
 
1) You are correct that the vertex is (0,0).

2)
the intensity of I light varies inversely as the sqiare of the distance D from the source. If the intensirty of illumination on a screen 5ft. from a light is 2 foot candles , what's the intensity on a screen 15 foot from the light?

Is the soultion 2/10 foot candle??

okay it's 2.9 got it...
Click to expand...

I = k*(1/d^2) = k/d^2

Substitute in the given info to solve for k:

2 = k/5^2
k = 50

Therefore, our eqn is I = 50/d^2.
Plug in the new distance (15 ft) and solve for I:

I = 50/15^2 = 50/225 = 2/9

I’m guessing when you wrote 2.9, you actually meant 2/9, but I just wanted to make sure.
 
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