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AleksAgata [21]

3 years ago

8

the area of a projected picture on a movie screen varies directly as the square of the distance from the projector to the screen

. if the distance of 20 feet produces a picture with an area of 64 square feet, what distance produces a picture with an area of 100 square feet?

1 answer:

Aleksandr [31]3 years ago

5 0

The answer is 1220 square feet

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Serggg [28]

Answer:

Step-by-step explanation:

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2 years ago

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LUCKY_DIMON [66]

For this problem all you have to do is plug in the value that they is in parenthesis for x. If it says g (x) = x and then it asks g(5) = ?, it is saying what happens if i put 5 in for every x. in this case it would be g (5) = 5. I just replaced x with 5.

So g (-2) we sub -2 for x
g (-2) = -2 (-2)^2 + 3 (-2) - 5
= -2 (4) - 6 - 5
= -19

g (0) = -2 (0) + 3 (0) - 5
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3 years ago

The scale on a map is 1 inch : 500 miles.

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3 years ago

Need help with #1 #4 #7 plz giving 15 points

valentina_108 [34]

Answer:

Q1: p = - 33

Q2: d = - 99

Q3: t = - 13

Step-by-step explanation:

Q1: $$ \textbf{-} \frac{\textbf{p}}{\textbf{3}} \hspace{1mm} \textbf{-} \hspace{1mm} \textbf{8} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{3}$

We solve this taking LCM.

We get: $\frac{-p - 24}{3} = 3$

$\implies - p - 24 = 9$

$\implies \textbf{p} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 33}$

Q4: $\frac{\textbf{d}}{\textbf{11}} \hspace{1mm} \textbf{-} \hspace{1mm} \textbf{4} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 13}$

Again we proceed like Q1 by taking LCM.

We get: $\frac{d - 44}{11} = - 13$

$\implies d - 44 = - 13 \times 11 = - 143$

$\implies d = - 143 + 44$

$\implies \textbf{d} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 99}$

Q7: 5t + 12 = 4t - 1

We club the like terms on either side.

$\implies 5t - 4t = - 1 - 12$

$\implies (5 -4)t = - 13$

$\implies \textbf{t} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 13}$

Hence, the answer.

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3 years ago

When working with inequalities what do you do when its x squared > 16 and vice versa does the sign flip.

S_A_V [24]

Answer:

$x4$

$\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}$

Step-by-step explanation:

$x^2>16$

$x\sqrt{16}$

$x4$

The interval notation will be:

$\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}$

The contrary,

$x^2$

is $-4$

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3 years ago

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