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Mrac [35]
2 years ago
10

What number is the opposite of -3? Explain your reasoning

Mathematics
1 answer:
Troyanec [42]2 years ago
3 0
The answer is 3

Answer 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
5/8 0.67 1.2 3/5 Greatest to least
SSSSS [86.1K]
1.2
0.67
5/8
3.5

I think this is what it would be.
4 0
3 years ago
No links or trolls please. I need help with this math, it’s due soon. Thanks you. It’s a division sign by the way, it’s hard to
xxTIMURxx [149]

Answer:

n^2p

Step-by-step explanation:

Thats the answer

3 0
3 years ago
What is the distance between the points (10, -7) and ( 10, 9) on the coordinate plane?
xenn [34]
16 on the y-axis, 0 on the x axis
7 0
3 years ago
Given points A (1, 2/3), B (x, -4/5), and C (-1/2, 4) determine the value of x such that all three points are collinear
AlladinOne [14]

Answer:

x=\frac{83}{50}

Step-by-step explanation:

we know that

If the three points are collinear

then

m_A_B=m_A_C

we have

A (1, 2/3), B (x, -4/5), and C (-1/2, 4)

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

step 1

Find the slope AB

we have

A(1,\frac{2}{3}),B(x,-\frac{4}{5})

substitute in the formula

m_A_B=\frac{-\frac{4}{5}-\frac{2}{3}}{x-1}

m_A_B=\frac{\frac{-12-10}{15}}{x-1}

m_A_B=-\frac{22}{15(x-1)}

step 2

Find the slope AC

we have

A(1,\frac{2}{3}),C(-\frac{1}{2},4)

substitute in the formula

m_A_C=\frac{4-\frac{2}{3}}{-\frac{1}{2}-1}

m_A_C=\frac{\frac{10}{3}}{-\frac{3}{2}}

m_A_C=-\frac{20}{9}

step 3

Equate the slopes

m_A_B=m_A_C

-\frac{22}{15(x-1)}=-\frac{20}{9}

solve for x

15(x-1)20=22(9)

300x-300=198

300x=198+300

300x=498

x=\frac{498}{300}

simplify

x=\frac{83}{50}

8 0
3 years ago
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