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3 years ago

16- X = -2 Solve for x

Mathematics

2 answers:

Licemer1 [7]3 years ago

7 0

Answer: x=18

Step-by-step explanation: 16-18=-2

Good luck! :)

shepuryov [24]3 years ago

4 0

Answer: -18

Step-by-step explanation: 16-x = -2

x = -16-2 [16 changes to -16 when transferred to RHS]

x = -18

Hope it helped u,

mark as the brainliest

^_^

You might be interested in

What are the first three terms of a geometric sequence in which as = 25 and the common ratio is 5?

marin [14]

Answer:

The first three terms are: 25, 125 and 625

Step-by-step explanation:

Given

$a =25; r = 5$

Required

The first three terms

The nth term of a geometry progression is:

$T_n =ar^{n-1$

So, we have:

First term: $n = 1$

$T_n =ar^{n-1$

$T_1 =25 * 5^{1-1$

$T_1 =25 * 5^0$

$T_1 =25 * 1$

$T_1 =25$

Second term: $n = 2$

$T_n =ar^{n-1$

$T_2 =25 * 5^{2-1$

$T_2 =25 * 5^{1$

$T_2 =25 * 5$

$T_2 =125$

Third term: $n = 3$

$T_n =ar^{n-1$

$T_3 =25 * 5^{3-1$

$T_3 =25 * 5^{2$

$T_3 =25 * 25$

$T_3 =625$

7 0

3 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