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2 years ago

Someone help please will mark brainliest

Mathematics

2 answers:

Arada [10]2 years ago

3 0

Answer:

the answer is 10

Step-by-step explanation:

sorry it took so long i had to do something

Shtirlitz [24]2 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Can some pls help me find the slope of this line i really really need help

Damm [24]

Answer:

Slope = Rise over Run

Rise = 2

Run = 3

Slope = 2/3

Let me know if this helps!

4 0

2 years ago

If 75% of the students in Toby’s grade voted, how many students are in Toby’s grade?

ELEN [110]

Answer:

The total number of students in Toby's grade is 200.

3 0

2 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

For every 1% increase in

givi [52]

Answer:

The GDP gap is 9 % when there is 4.5 % unemployment.

Step-by-step explanation:

The statement shows a reverse relationship, where an increase in unemployment is following by decrease in potential GDP and can be translated into the following rate:

$r = \frac{2\,\% \,GDP}{1\,\% unemp.}$

The GDP gap at a given increase in unemployment can be estimated by the following expression:

$\frac{g}{u} = r$

$g = r\cdot u$

Where:

$r$ - GDP gap-unemployment increase rate, dimensionless.

$u$ - Increase in unemployment rate, measured in percentage.

$g$ - GDP gap, measured in percentage.

If $r = \frac{2\,\% \,GDP}{1\,\% unemp.}$ and $u = 4.5\,\%\,unemp.$ , the GDP gap is:

$g = \left(\frac{2\,\%\,GDP}{1\,\%\,unemp.} \right)\cdot (4.5\,\%\,unemp.)$

$g = 9\,\%\,GDP$

The GDP gap is 9 % when there is 4.5 % unemployment.

3 0

3 years ago

An average person in the United States throws away 4.5 × 10² g of trash per day. In 2013, there were about 3.2×1063.2×106 people

Mashcka [7]

Well, Since every 1 person throws away 4.5 × 10² g trash each day (365 days in each year), You would times 3.2×1063.2×106 by the 4.5 × 10² g to get how much trash was thrown away each day and then times that by 365 to figure how much trash was thrown away the entire year

8 0

2 years ago