OH LORD THIS IS A NIGHTMARE FOR ME

Mathematics

1 answer:

monitta3 years ago

7 0

Answer:

∠1 = 150°

Step-by-step explanation:

supplementary angle to angle 1 = 180° - 100° - 50° = 30°

∠1 = 180° - 30° = 150°

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How do I solve this

Kitty [74]

Hello,

$x^2-2x-27=-5x+1\\

x^2-2x+5x-27-1=0\\

x^2+3x-28=0\\

x^2+7x-4x-28=0\\

x(x+7)-4(x+7)=0\\

(x+7)(x-4)=0\\

x=-7\ or\ x=4\\
$

8 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D

erik [133]

2^3
——
3^3

............

8 0

2 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