Divide 13 by v. Then, subtract 7.

Mathematics

2 answers:

Arte-miy333 [17]4 years ago

8 0

Answer:

b........

Step-by-step explanation:

g100num [7]4 years ago

3 0

Correct answer is C, you would do the order of operation 13\v then take that and subtract 7

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What is 2/3 – (-2/3)<br> ?A.-4/3<br> B.4/3<br> C.0<br> D.2/3

dalvyx [7]

Answer:

It's C, they both cancel each other out

6 0

3 years ago

Read 2 more answers

How do I solve this?

beks73 [17]

Distribute the exponent power of 2 to each term:
[2² * (√(3x⁵))² - 2² * (√y)²]

Now simplify. Placing a square root to the power of 2 just cancels each other out so you're left with:
[4(3x⁵) - 4y]

Simplify again for your answer:
12x⁵ - 4y

6 0

4 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1