All categories

3 years ago

BRAINLIEST!!!!! PLEASE ANSWER!!!!!

Mathematics

1 answer:

Natalka [10]3 years ago

5 0

Answer:

11/5

Step-by-step explanation:

the line goes up 11, and over 5, therfore the slope would be rise over run...so 11/5

You might be interested in

Ray SP bisects mZRST.<br> If m RSP = 3X – 2 and m PST = 9x -26<br> Answer: Find : mZRST

ludmilkaskok [199]

Answer:

RST = 20

Step-by-step explanation:

a ray that divides an angle into two equal angles is called an angle bisector.

So RSP = PST

3 x -2 = 9x - 26

6x = 24

x = 4

RST = 3x -2 + 9x -26 = 20

5 0

3 years ago

Find the possible values of r in the inequality 5 > r - 3.

ioda

The answer to this question is A

3 0

4 years ago

Read 2 more answers

How do I solve this question?

FrozenT [24]

18
Do u give brainiest?

8 0

3 years ago

Can some one help me with 1,3,and 5

FrozenT [24]

1. 2/100 or 1/50, as 0.02 is equal to 2 hundreths. Simplifying gets you 1/50

3. 1/2 or 50/100, as 0.50 is equal to 0.5 which is equal to 50 hundreths. Simplify to get 1/2

5. 88/100 or 44/50 , as 0.88 is equal to 88 hundreths. After simplifying, you get 44/50

Depending on your teacher, you may be told to not simplify. It depends

8 0

3 years ago

Integrate dx/3sinx+4cosx

german

$\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}$

A standard approach would be the tangent half-angle substitution:

$t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx$

Then

$\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}$

$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}$

from which we get

$\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt$

So the integral becomes

$\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}$

Rewrite the denominator as

$2t^2-3t-2=(2t+1)(t-2)$

and expand the integrand into its partial fractions:

$\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)$

We have

$\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C$

$=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C$

$=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C$

6 0

3 years ago