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Contact [7]
3 years ago
6

1) If y=4 when x= -2, find x when y= -5

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0

1) x = 2.5 when y = -5

2) x = 0.25 when y = - 1

<u>Solution:</u>

Given that , we have to use direction variation formula.

Now, we know that, direct variation formula is y = kx, where k is proportionality constant.

1) If y=4 when x= -2, find x when y= -5  

Put y = 4 and x = -2 in y = kx  

4 = k(-2)  

k = - 2

Then, when y = -5  

-5 = (- 2)x  

x = 5/2  

x = 2.5

2) If y=20 when x = -5, find x when y= -1

Put y = 20 and x = -5 in y = kx  

20 = k(-5)  

k = - 4

Then, when y = -1  

-1 = (- 4)x  

x = 1/4  

x = 0.25

Hence, 1) x = 2.5 when y = -5 and 2) x = 0.25 when y = - 1

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Anna35 [415]

Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.

Given the function is f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}  and the Rolle's Theorem does not apply to the function.

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The condition for Rolle's theorem to be true as:

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  • f(x) must be continuous in [a,b].
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To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.

If we look closely at the given function we can see that the first derivative of the given function is:

\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end

From this point of view we can see that the given function is not defined for x=0.

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Learn more about Rolle's Theorem from here brainly.com/question/12279222

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8 0
2 years ago
What is $5.60 for 4 pounds as a unit rate.
Sphinxa [80]

so you can either do pounds per dollar or dollars per pound:

pounds per dollar : about 0.71 lbs / $

dollars per pound : $1.40 / lb

i hope this helped, based on the question i believe that it is dollars per pound but its good to know both and have a nice day


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If A/B and C/D<br> are rational expressions, then<br> AxC/BxD = AxD/BxC<br> A. True<br> B. False
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Answer:

True

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3 years ago
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Otrada [13]

Answer:

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