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Andrei [34K]
3 years ago
6

For this question, f(x) = 4kx2 + (4k + 2)x + 1, where k is a real constant.

Mathematics
1 answer:
snow_tiger [21]3 years ago
6 0

Answer:

fghhjhhfgkus+($+#($!?$)

You might be interested in
What is the argument of -5\sqrt{3}+ 5i?
Shtirlitz [24]

Answer:

150 degrees

Step-by-step explanation:

Graphing the complex number we see the angle terminates in the second quadrant. This means the argument, the angle, will be between 90 degrees and 180 degrees.

So if we create a right triangle with that point after graphing it. We see the height of that triangle is 5 because that is the imaginary part. The base of that triangle has length 5\sqrt{3}. The problem is this doesn't give us any part of the angle we want, but it does give us the complementary of the part of the angle that is in second quadrant.

Let's find the complementary angle.

So the opposite side of the complementary angle is 5.

The adjacent side of the complementary angle is 5\sqrt{3}.

\tan(\theta)=\frac{5}{5\sqrt{3}}

\tan(\theta)=\frac{1}{\sqrt{3}}

\theta=\tan^{-1}(\frac{1}{\sqrt{3}})

\theta=30

So 90-30=60.

The answer therefore 60+90=150.

4 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

6 0
3 years ago
Rewrite the following equation in standard form.<br> y + 4 = 5(x - 8)
Helen [10]

Answer:

5x-y-36=0

Step-by-step explanation:

y+4=5x-40

5x-y=36

4 0
3 years ago
What is the value of x in the product of powers below?
sashaice [31]

Mmm, x is equals to -7....

8 0
3 years ago
4 = 1 - n + 4? Geometry btw
eimsori [14]

Answer:

N = -7

Step-by-step explanation:

4 = 1 - n -4

Subtract the numbers first

1 - 4 = -3

Then rearrange the terms

4 = -3 - n

-----> 4 = n - -3

The you add 3 to both sides of the equations

4 + 3 = -n - 3 + 3

3 0
2 years ago
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