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3 years ago

Y=2x^2-2x+7 in vertex form

Mathematics

1 answer:

telo118 [61]3 years ago

7 0

Hello,

y=2x²-2x+7
==>y=2(x²-2*1/2x+1/4)=7-1/2
==>y=2(x-1/2)²+13/2

Vertex=(1/2,13/2)

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3 years ago

What is 4 to the power of x minus 3 =18

mote1985 [20]

3 to the power of 4:

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2 years ago

What is the value of x3 × y1 when x = 5 and y = 10

katovenus [111]

Answer:

150

Step-by-step explanation:

To solve the problem, put 5 instead of x, so 5*3

and put 10 instead of y, so 10*1

when combined,

(5*3) * (10*1)

= 15 * 10

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Hope this helps!!

Let me know if I'm wrong...

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2 years ago

Choose the answers that have a repeating decimal?

Dmitry [639]

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3 years ago

If f(x) = 9x10 tan−1x, find f '(x).

djverab [1.8K]

Answer:

$\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = 9x^{10} \tan^{-1}(x)$

Step 2: Differentiate

[Function] Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Basic Power Rule: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Arctrig Derivative: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

2 years ago