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Ilia_Sergeevich [38]

2 years ago

10

A rectangular picture measuring 5 in. by 9 in. is surrounded by a frame with uniform width x. Write a quadratic function in stan

dard form to show the combined area of the picture and fram

2 answers:

disa [49]2 years ago

6 0

A=(5+2x)(9+2x)

A=45+10x+18x+4x^2

A=4x^2+28x+45

dedylja [7]2 years ago

4 0

(x.x)-(5x9)= Combined area of the picture and frame.

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c Aerobics and jumping rope

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2 years ago

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ale4655 [162]

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3 years ago

Read 2 more answers

An SRS of size 20 is drawn from a population that has a normal distribution. The sample has a mean of 122.5 and a standard devia

Tems11 [23]

Answer:

The value is $\sigma_{\= x} = 1.789$

Step-by-step explanation:

From the question we are told that

The sample size is n = 20

The sample mean is $\= x = 122.5$

The standard deviation is $s = 8$

Generally the standard error of the mean is mathematically represented as

$\sigma_{\= x} = \frac{s}{\sqrt{n} }$

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8 0

2 years ago

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

ivanzaharov [21]

$\Large \mathbb{SOLUTION:}$

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

5 0

2 years ago

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

charle [14.2K]

ANSWER

See below

EXPLANATION

Given

$f(x) = \frac{ {x}- 9 }{x + 5}$

and

$g(x) = \frac{ - 5x - 9}{x - 1}$

$(f \circ \: g)(x)= \frac{ (\frac{ - 5x - 9}{x - 1})- 9 }{(\frac{ - 5x - 9}{x - 1} )+ 5}$

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9 - 9(x - 1)}{x - 1}}{\frac{ - 5x - 9 + 5(x - 1)}{x - 1} }$

Expand:

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9 - 9x + 9}{x - 1}}{\frac{ - 5x - 9 + 5x - 5}{x - 1} }$

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9x + 9 - 9}{x - 1}}{\frac{ - 5x + 5x - 5 - 9}{x - 1} }$

$(f \circ \: g)(x)= \frac{ \frac{ - 14x }{x - 1}}{\frac{ -14}{x - 1} }$

Since the denominators are the same, they will cancel out,

$(f \circ \: g)(x)= \frac{ - 14x}{ - 14} = x$

8 0

2 years ago