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MrRa [10]
3 years ago
5

Ax) = x² + 4x + 5 What is the vertex?

Mathematics
1 answer:
kakasveta [241]3 years ago
6 0

Answer:

(-2,1)

Step-by-step explanation:

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What is a cubic polynomial function in standard form with zeros –3, 2, and–3?
levacccp [35]

Answer:

(x-2)(x+3)(x+3)=0\\

Step-by-step explanation:

Multiply those together and combine like terms.  Then make sure you order the terms in descending powers.

7 0
3 years ago
What is the slope, y-intercept and slope intercept equation to this graph. Plz help
fredd [130]

Answer:

y intercept is 3 and slope is 4/1

Step-by-step explanation:

3 0
3 years ago
Use zero property to solve the equation.<br> F(x)=3x(x+7)-2(x+7)
oksian1 [2.3K]

Answer:

x = -7 or x = 2/3

Step-by-step explanation:

I'm assuming you meant solve for x when f(x) = 0.

f(x) = 3x(x + 7) - 2(x + 7)

0 = (x + 7)(3x - 2) -- Both terms have a common factor of (x + 7) so we can group them

x + 7 = 0 or 3x - 2 = 0 -- Use ZPP

x = -7 or x = 2/3 -- Solve

7 0
3 years ago
Help me with just 27​
tankabanditka [31]

Answer:

16/13

Step-by-step explanation:

the error is when it was divided, one of the fractions should be reversed to be multiplied

<u> </u><u>4</u><u> </u><u> </u> ÷ <u> </u><u>1</u><u>3</u><u> </u>

7. 28

<u> </u><u>4</u><u> </u> × <u>2</u><u>8</u><u> </u> ,like this

7. 13

= <u> </u><u>1</u><u>1</u><u>2</u><u> </u> ,divide both by 7

91

= <u> </u><u>1</u><u>6</u><u> </u>

13

7 0
3 years ago
find the Taylor Series for tan−1(x) based at 0. Give your answer using summation notation and give the largest open interval on
enyata [817]

Let f(x)=\tan^{-1}x. Then f'(x)=\frac1{1+x^2}. Note that f(0)=0.

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n

This means that for |-x^2|=|x|^2, or -1, we have

\displaystyle f'(x)=\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n\ge0}(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}

Integrate the series to get

f(x)=f(0)+\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}

\implies\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}

6 0
3 years ago
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