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zhenek [66]
3 years ago
14

PLS HELP ME THIS IS HARDD

Mathematics
1 answer:
pogonyaev3 years ago
6 0

Answer:

it would be the second answer because the shape has to get 3/4 smaller than the pre-image

Step-by-step explanation:

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**Spam answers will not be tolerated**
Morgarella [4.7K]

Answer:

f'(x)=-\frac{2}{x^\frac{3}{2}}

Step-by-step explanation:

So we have the function:

f(x)=\frac{4}{\sqrt x}

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Therefore, our derivative would be:

\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}

Place the 4 in front:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})

Distribute:

=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}

The numerator will use the difference of two squares. Thus:

=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Simplify the numerator:

=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Both the numerator and denominator have a h. Cancel them:

=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Now, substitute 0 for h. So:

=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})

Simplify:

=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

=4( \frac{-1}{(x)(2\sqrt{x})})

Multiply across:

= \frac{-4}{(2x\sqrt{x})}

Reduce. Change √x to x^(1/2). So:

=-\frac{2}{x(x^{\frac{1}{2}})}

Add the exponents:

=-\frac{2}{x^\frac{3}{2}}

And we're done!

f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}

5 0
3 years ago
What is the percent decrease from 376 to 0.
Alekssandra [29.7K]

hey its -100 percent. the negative to show decrease.

6 0
2 years ago
Read 2 more answers
A department store has a discount box of cell phone cases. In the box are 10 leather cases, 12 protective plastic cases, and 6 f
Kisachek [45]
28 cell phone cases in all. 
6 of the 28 cases are fabric. 

6/28 simplified to 3/14 

So the answer to your question would be 3/14

Hope this helps! :) 
7 0
3 years ago
Read 2 more answers
What is the probability that a 44% free throw shooter will miss her next free throw?
stellarik [79]
The probability is 56/100 or 56%. Because they only got 44% of their previous throws, the probability that they'll miss the next one is 56%.
3 0
3 years ago
What are the real zeros of the function g(x) = x3 + 2x2 − x − 2?
Nat2105 [25]
Upon a slight rearrangement this problem gets a lot simpler to see.

x^3-x+2x^2-2=0  now factor 1st and 2nd pair of terms...

x(x^2-1)+2(x^2-1)=0

(x+2)(x^2-1)=0  now the second factor is a "difference of square" of the form:

(a^2-b^2) which always factors to (a+b)(a-b), in this case:

(x+2)(x+1)(x-1)=0

So g(x) has three real zero when x={-2, -1, 1}
4 0
2 years ago
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