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3 years ago

Plz help i beg uu. Will give branliest

Mathematics

2 answers:

galben [10]3 years ago

6 0

The answer is C hope this helps!!

icang [17]3 years ago

4 0

I think the answer would be C :$:$:$:$:$:$:$:$:&:$

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What is the midpoint between -2-3i and 3+9i

Svetlanka [38]

Answer:

1/2 + 3i is the midpoint between -2-3i and 3+9i.

Step-by-step explanation:

Given the complex number

-2-3i

3+9i

The formula to find the midpoint of two complex number (a + bi) and (c + di) is:

$M=\frac{\left(a+c\right)}{2}+\frac{\left(b+d\right)i}{2}$

$M=\frac{\left(-2+3\right)}{2}+\frac{\left(-3+\left(9\right)\right)i}{2}$

$M=\frac{-2+3+\left(-3+9\right)i}{2}$

$M=\frac{1+6i}{2}$

$M=\frac{1}{2}+3i$

Therefore, 1/2 + 3i is the midpoint between -2-3i and 3+9i.

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3 years ago

Right to explain how this digit 6 changes value in number 666, 666, 666

zvonat [6]

To a higher number the six changes value in this number above

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3 years ago

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$