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3 years ago

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1 answer:

stepladder [879]3 years ago

6 0

Set R(x) equal to C(x). Solve the resulting equation for x, paying attention to the domain of C(x):

Solve 200x-2x^2 = -x^2 + 25x + 31000 for x greater than or equal to zero but less than or equal to x. That means you must reject any solution outside of this domain.

Rearrange all of these terms on the left side of your equation and place zero on the right side. Next, rearrange all of the terms on the left side in descending order by powers of x. Next, solve the resulting quadratic equation. Make certain to reject any x that does not fall within [0,100].

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What is center: (-4,-12) Radius: 4 ?

netineya [11]

Answer:

have a great day ❤️❤️☺️

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2 years ago

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The length and width of a rectangle are in a 3:5 ratio. The perimeter of the rectangle is 64. What are the length and width of t

Levart [38]

Hey there! I'm happy to help!

We know that the perimeter is 2L+2W. Le'ts plug in the values 3 and 5 from our ratio and see what our perimeter would be with that.

2(3)+2(5)

6+10

16

How can we get this 16 to 64?

64/16=4

So, we need to multiply this length and width by four to get 64 as our perimeter!

3*4=12

5*4=20

We can plug this into our perimeter equation

2(12)+2(20)

24+40

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Therefore, our length is 12 and our width is 20.

Have a wonderful day! :D

5 0

2 years ago

If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?

svetoff [14.1K]

$sin2\alpha = \frac{260}{269}$

<u>Step-by-step explanation:</u>

We have , $Tan\alpha = \frac{Perpendicular}{Base} = \frac{10}{13}$ ,

We know that $sin\alpha = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }$

Substituting values of P & B , $sin\alpha = \frac{10}{\sqrt{10^{2} + 13^{2}} } = \frac{10}{\sqrt{269} }$

Now , $sin2\alpha = 2sin\alpha cos\alpha = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }$

⇒ $sin2\alpha =$ $\frac{10}{\sqrt{269} }$ × $\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }$ ×2

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} } )$

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} } )$

⇒ $sin2\alpha = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} } )$

⇒ $sin2\alpha = \frac{260}{269}$

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2 years ago

A welder works 8 hours a day at a rate of $25.00 per hour. The overtime rate is doubled the normal rate. On a particularday, the

Alborosie

Answer:

C.

Step-by-step explanation:

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2 years ago

ellen had 25 hair clips,7 of them were blue and the rest are purple.what percent of hair clips were purple

strojnjashka [21]

Answer=72%

25-7=18purple clips

18 x
____=____
25 100

cross multiply
25x=1800
divide both sides by 25
x=72%

6 0

3 years ago

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