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3 years ago

Multiply (x + 1); x + 1 3+ + 2x+2 x +x+1 2+2x+1

Mathematics

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

HI!!! I believe your answer is option a because it looks like your not multiplying anything so hope this helps:)

Step-by-step explanation:

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Exponential function

laiz [17]

A typical exponential function is y=a $b^{x}$
when x=3.5, y=16.2
when x=6, y=3936.6
plug these values into the exponential function:
16.2=a $b^{3.5}$
3936.6=a $b^{6}$
divide the second equation by the first to eliminate a:
243= $b^{(6-3.5)}$
log both sides: log243=2.5logb
logb=log243/2.5
use your calculator to find b: b=3.6
plug b=3.6 in the first equation to find a:
16.2=a $* (3.6)^{3.5}$
a=0.183

please double check my calculation

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marusya05 [52]

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Step-by-step explanation:

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3 years ago

How to solve this?<br>\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x

ivanzaharov [21]

$\Large \mathbb{SOLUTION:}$

$\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes} \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}$

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3 years ago