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3 years ago

Help help help help help please

Mathematics

1 answer:

iris [78.8K]3 years ago

4 0

Answer:

Step-by-step explanation:

h + 4g

5 + 4(8)

5 + 32 = 37

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Please answer quickly thanks

Anastaziya [24]

8/3 = 2 2/3 and -11/3 = -3 2/3
so the middle line is 0 and each big line is 1.
Put the first dot on 2 2/3 or right before the 3rd big line.
Put the second dot on -3 2/3 or the line right after the 4th big line going back from 0

3 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

For what interval is the function f(x) = (20 + \sqrt{x}) / (\sqrt{20 + x}) continuous?

konstantin123 [22]

Try this solution:
1. according to the condition
$\left \{ {{x \geq 0} \atop {20+x\ \textgreater \ 0}} \right. \ =\ \textgreater \ \ x \geq 0.$
2. for more details see the attached graph.

Answer: [0;+oo)

5 0

3 years ago

if it asks for scale factor does that mean i’m supposed to write a ratio? Can someone solve and explain these for me i’m so conf

STatiana [176]

Yes scale factor means ratio. The height of the smaller pyramid to the larger pyramid in ratio form is 8:24 which simplifies to 1:3. The ratio of its areas is the scale factor squared, so 1^2:3^2=1:9. Since the surface area of the smaller pyramid is 124, the surface area of the larger pyramid must be 124x9. The ratio of volumes is the scale factor cubed, so 1^3:3^3=1:27. If the volume of the larger pyramid is 648, then the volume of the smaller pyramid is 1/27 of that, which equals 24.

8 0

3 years ago

Olivia and Camille went clothes shopping. Camille spent $19.40 less than twice what Olivia spent. In total, they spent $272.35.

nydimaria [60]

if Camille spent 19.40$ LESS than DOUBLE what Olivia spent then a rough estimate already can be Olivia spent 100 and Camille spent 172 but to find this out we have to add 19.40 onto 272.35 = 291.75 then we now that 291.75 is thrice what Olivia spent so if we third it we know what Olivia spent so 291.75/3 = 97.25 leaving 194.50 then if we minus 19.40 off 194.50 we get 175.10 so:

Camille spent: 175.10 and

Olivia Spent: 97.25

4 0

3 years ago