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3 years ago

Simplify -4b+8c+12-8b-2c+6

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

5 0

Answer:

-4b+8c+12-8b-2c+6

-4b-8b+8c-2c+12+6

-12b+6c+18.

Thank you.

BY GERALD GREAT.

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John bought new headphones originally listed for $ 80.99 . They are \(35\%)| off. Which equation can be used to find the amount

Katarina [22]

Answer:

$28.35

Step-by-step explanation:

The headphones are $80.99 and are 35% off. To find the amount he will save, change the 35% to a decimal by moving it two times to the left, which is 0.35. Multiply 80.99 and 0.35 to get 28.3465. Round to get $28.35.

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2 years ago

During which time period does Landon's elevation change the fastest? Explain how you know?

bogdanovich [222]

<h3>1. How many inches per minute does London's elevation change between 4 minutes and 8 minutes. </h3>

The question actually asks for the slope of the line that stands for the points $(4,3) \ and \ (8,6)$ why? because the questions tells us that London's elevation changes between 4 minutes and 8 minutes here. Hence, to find the slope of this line we have to use the following formula:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(4,3) \\ P(x_{2},y_{2})=P(8,6)$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(4,3) \\ P(x_{2},y_{2})=P(8,6) \\ \\ So: \\ \\ m=\frac{6-3}{8-4}=0.75in/min$

<em>So London's elevation changes 0.75 inches per minute</em>

<h3>2. During which time period does London's elevation change the fastest?</h3>

The greater the absolute value of the slope of the line the faster London's elevation changes. Since this is a Piecewise function, we must analyze each period.

FIRST:

→ Between 0 minutes and 4 minutes the function is constant, so there is no any change here.

→ Between 10 minutes and 14 minutes the function is constant, so there is no any change here.

→ Between 18 minutes and 22 minutes the function is constant, so there is no any change here.

So the solution is not in these parts of the function.

SECOND:

→ Between 4 minutes and 10 minutes the function has a positive slope, so there is change here.

In the previous item we calculated the slope between 4 and 8 minutes that is the same slope between 4 and 8 minutes and equals 0.75.

→ Between 14 minutes and 18 minutes the function has a positive slope, so there is change here.

Let's take two points here, say, $(16,5) \ and \ (18,3)$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ But: \\ \\ P(x_{1},y_{1})=P(16,5) \\ P(x_{2},y_{2})=P(18,3) \\ \\ So: \\ \\ m=\frac{3-5}{18-16}=-1 in/min$

As you can see, the absolute value here is 1 that is greater than 0.75.

<em>In conclusion, London's elevation changes the fastest between 14 and 18 minutes</em>

7 0

3 years ago

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

3 years ago

Which one is it A. B. C. or D.

san4es73 [151]

The answer will be c
Hope it's help you

6 0

3 years ago

Write and evaluate an expression for: Music lessons cost $20 a week. How much do 6 weeks of lessons cost?

givi [52]

Answer:

$120 dollars.

Step-by-step explanation: