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3 years ago

I have no idea how to do this. Please help.

Mathematics

1 answer:

ioda3 years ago

6 0

Answer:

C. 14,871.21

Step-by-step explanation: You take the number 29,742.42 and multiply it by .2 to get how much they spend on food (5,948.48), and you multiply the same number by .3 to get how much they spend on housing (8,922.73). Then, you add them up to get the total. .2 is acctually 20% of what they spend out of the 29,742.42 and the same for the .3. Hope this wasn't too complicated by the way I explained it.

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What is the best estimate for the percent of students scoring greater than 92 on at test?

son4ous [18]

Answer:

Step-by-step explanation:

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3 years ago

Translate the following sentence into math symbols. Then solve the problem. Show your work and keep the equation balanced.

adell [148]

For this case we have to represent the following expression algebraically:

"10 less than x is -45"

So:

10 less than x is represented as: $x-10$

Then, the complete expression is represented as:

$x-10 = -45$

Adding 10 to both sides of the equation:

$x = -45 + 10\\x = -35$

Thus, the value of the variable is $x = -35$

Answer:

$x = -35$

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3 years ago

John purchased a carr in 1961 for $12,000. Experts estimate that its value will increase by 5% per year. How much will be the co

GuDViN [60]

Answer:

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Step-by-step explanation:

6 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Help please. Seriously. I've posted this question 2 times. I seriously need help on this, guys.

Svetradugi [14.3K]

Answer:

20s=m

Step-by-step explanation:

I really don't understand it that well but all that I can is that s can represent students that went on the trip, and each student's ticket cost is 20$, so 20*s=20s, but we are taking m as the total money collected, so we can end up saying that 20s=m, hope u understood :)

8 0

3 years ago

Read 2 more answers