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3 years ago

Find the value of x for the question shown above

Mathematics

1 answer:

Sindrei [870]3 years ago

8 0

hint; the sum of all angles in a triangle is 180° !

^so all together it’s gonna be 180 !

hope this helped !

i’m pretty sure the value of it (x) is 12 but am not sure

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M +9 = -3<br> 8<br> What is the answer for m

harina [27]

Answer:

m=-12

Step-by-step explanation:

Not sure where the 8 comes from, but you would subtract 9 from the left side. The result is -12.

3 0

3 years ago

-2y+16*4<br> plz answer asap

Sedaia [141]

-2(y-32) i am 99% sure

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

Can someone please help Simple the square root 24+ square root 25

earnstyle [38]

Answer:

5+(2)*(sq root 6)

Step-by-step explanation:

This is pretty simple, a given is that the square root of 25 is 5, so you already have one of your numbers. Since the square root of 24 is not a perfect square, it must be reduced and then added separately to 5. The sq root of 24 is reduced to 2 multiplied by the square root of 6.

4 0

3 years ago

Read 2 more answers

How would you identify the number of real roots of a polynomial function from the graph of the function

beks73 [17]

Hi there! See an attachment below! (May not be clear for an explanation so I will explain it to you.)

We can find real roots of polynomial function by counting their x-intercepts of graph. See in the attachment that the graph intersects 3 points on x-axis/plane.

x-axis represents the solutions of the graph or equation of the graph. For example, a graph of (x-1)² would be a parabola that has x-intercept at (1,0). If we solve the equation for (x-1)² = 0, our solution would be x = 1 like the x-intercepts.

Conclusions

We can find the real roots of the graph by counting how many points does the graph intersect on x-axis/plane.
If a graph doesn't intersect x-axis or does not have any points on x-axis, that graph or its equation does not have a real root. For example, graph of x²+1

Any questions can be asked through comment as I may reply to you soon! Thanks for using Brainly. Have a great day and happy learning!

3 0

3 years ago

Find the value of x for the question shown above​

Find the value of x for the question shown above