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2 years ago

Pls help i dunno what to do with these problems

Mathematics

2 answers:

Mariulka [41]2 years ago

8 0

I Think it’s D sorry if I’m wrong yo

Evgesh-ka [11]2 years ago

3 0

Answer: D

Step-by-step explanation:

I've done this

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Chang is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices.

Elena L [17]

Answer:

Step-by-step explanation:

86-65=21$ (initial fee diffrence)

21/0.6= 35

for 35 m company B charge less than A unless if renter want to drive further 35m so company A present a better offer.

6 0

3 years ago

Wha is 5÷312 in long division

bezimeni [28]

The answer is 62 with a remainder of 2.

Hope this helps!

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3 years ago

Which parent function is represented by the graph?

ikadub [295]

The parent function represented by the graph is an exponential function, as it gets exponentially smaller.

8 0

3 years ago

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jackie made bracelets for 8 days. when he was done he had 96 bracelets. write and equation to express how many bracelets jackie

julsineya [31]

96 divided by 8.
Answers 12 bracelets per day

4 0

3 years ago

Read 2 more answers

Consider the function.

Verdich [7]

It looks like you're given

$f(x) = 1012x^{101} - 72x^{75} + \pi x^2 - e^{2x} + 100346$

and are asked to find the 102nd derivative of f(x).

Recall the power rule: for integer n,

$\displaystyle \left(x^n\right)' = nx^{n-1}$

This means that the power of x reduces to 0 after differentiating n times, and you're left with a constant coefficient n! :

• after differentiating 2 times,

$\left(x^n\right)'' = \left(nx^{n-1}\right)' = n(n-1)x^{n-2}$

• after differentiating 3 times,

$\left(x^n\right)^{(3)} = \left(n(n-1)x^{n-2}\right)' = n(n-1)(n-2)x^{n-3}$

• and so on, up to the n-th time, which yields

$\left(x^n\right)^{(n)} = n(n-1)(n-2)\cdots\times2\times1x^{n-n} = n!$

As soon as you have a constant, the next derivative will be 0. This means that after differentiating 102 times, the first 3 terms of f(x), as well as the constant term, will vanish.

Recall the chain rule:

$\bigg(f(g(x))\bigg)' = f'(g(x)) \times g'(x)$

Then the first few derivatives of the exponential term are

$\left(e^{2x}\right)' = e^{2x} \times (2x)' = 2e^{2x}$

$\left(e^{2x}\right)'' = 2\left(e^{2x}\right)' = 2^2e^{2x}$

$\left(e^{2x}\right)^{(3)} = 2^2\left(e^{2x}\right)' = 2^3e^{2x}$

and so on, with n-th derivative

$\left(e^{2x}\right)^{(n)} = 2^ne^{2x}$

Putting everything together, we have

$\boxed{f^{(102)}(x) = -2^{102}e^{2x}}$

6 0

3 years ago