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den301095 [7]
2 years ago
6

A searchlight uses a parabolic mirror to cast a beam of light in parallel rays where the light bulb is located at the focus of t

he parabola in order to give the best illumination. The mirror is modeled by y^2=36(x-10), where the measurements are in cm. What is the location of the light bulb?

Mathematics
1 answer:
Vitek1552 [10]2 years ago
3 0

Answer:

the third option

Step-by-step explanation:

Our equation is

{y}^{2}  = 36(x - 10)

Equation of a vertical parabola is

( {y - k)}^{2}  = 4p(x  - h)

where (h,k) is the center

The focus is

(h+p, k)

Equation of directrix is

x= h-p,

Here the center is (10,0)

Next, we factor out 4 in the original equation

4(9)

So we have

{y}^{2}  = 4(9)(x - 10)

So our p=9,

So our focus is

(10+9,0) or (19,0)

The third option is the answer

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10+8+7(-10)-(-1)
Elina [12.6K]

Answer:

-51

Step-by-step explanation:

Let's do order of operations.

10+8-70-(-1)     <- We did negative 7 times 10.

18-70-(-1)    <- We did 10 plus 18.

-52-(-1)     <- We did 18 minus 70

When you open the () we get:

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5 0
3 years ago
-X-5y + z = 17
grin007 [14]

Answer:

x = -1 , y = -4 , z = -4

Step-by-step explanation:

Solve the following system:

{-x - 5 y + z = 17 | (equation 1)

-5 x - 5 y + 5 z = 5 | (equation 2)

2 x + 5 y - 3 z = -10 | (equation 3)

Swap equation 1 with equation 2:

{-(5 x) - 5 y + 5 z = 5 | (equation 1)

-x - 5 y + z = 17 | (equation 2)

2 x + 5 y - 3 z = -10 | (equation 3)

Subtract 1/5 × (equation 1) from equation 2:

{-(5 x) - 5 y + 5 z = 5 | (equation 1)

0 x - 4 y+0 z = 16 | (equation 2)

2 x + 5 y - 3 z = -10 | (equation 3)

Divide equation 1 by 5:

{-x - y + z = 1 | (equation 1)

0 x - 4 y+0 z = 16 | (equation 2)

2 x + 5 y - 3 z = -10 | (equation 3)

Divide equation 2 by 4:

{-x - y + z = 1 | (equation 1)

0 x - y+0 z = 4 | (equation 2)

2 x + 5 y - 3 z = -10 | (equation 3)

Add 2 × (equation 1) to equation 3:

{-x - y + z = 1 | (equation 1)

0 x - y+0 z = 4 | (equation 2)

0 x+3 y - z = -8 | (equation 3)

Swap equation 2 with equation 3:

{-x - y + z = 1 | (equation 1)

0 x+3 y - z = -8 | (equation 2)

0 x - y+0 z = 4 | (equation 3)

Add 1/3 × (equation 2) to equation 3:

{-x - y + z = 1 | (equation 1)

0 x+3 y - z = -8 | (equation 2)

0 x+0 y - z/3 = 4/3 | (equation 3)

Multiply equation 3 by 3:

{-x - y + z = 1 | (equation 1)

0 x+3 y - z = -8 | (equation 2)

0 x+0 y - z = 4 | (equation 3)

Multiply equation 3 by -1:

{-x - y + z = 1 | (equation 1)

0 x+3 y - z = -8 | (equation 2)

0 x+0 y+z = -4 | (equation 3)

Add equation 3 to equation 2:

{-x - y + z = 1 | (equation 1)

0 x+3 y+0 z = -12 | (equation 2)

0 x+0 y+z = -4 | (equation 3)

Divide equation 2 by 3:

{-x - y + z = 1 | (equation 1)

0 x+y+0 z = -4 | (equation 2)

0 x+0 y+z = -4 | (equation 3)

Add equation 2 to equation 1:

{-x + 0 y+z = -3 | (equation 1)

0 x+y+0 z = -4 | (equation 2)

0 x+0 y+z = -4 | (equation 3)

Subtract equation 3 from equation 1:

{-x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = -4 | (equation 2)

0 x+0 y+z = -4 | (equation 3)

Multiply equation 1 by -1:

{x+0 y+0 z = -1 | (equation 1)

0 x+y+0 z = -4 | (equation 2)

0 x+0 y+z = -4 | (equation 3)

Collect results:

Answer: {x = -1 , y = -4 , z = -4

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