Ask question

Ask question

All categories

andrey2020 [161]

3 years ago

6

URGENT!!!!!!!!! 40 POINTS AND BRAINLIEST!!!!!!!!!!!!!

2 answers:

faltersainse [42]3 years ago

8 0

Answer:

the right awnser is 75 and plzz try to be calm it'll be okay trust me.

Step-by-step explanation:

cricket20 [7]3 years ago

3 0

Answer: 75 degrees

Step-by-step explanation:

You might be interested in

What is 56cm added to 2m 50cm?

Over [174]

100cm = 1m so 2m 50cm = 250cm + 56cm= 306cm which means the answer is 3m 6 cm

3 0

4 years ago

Please answer this question above.

ioda

The distance between point J and line KL would be the line perpendicular to the line, which is line JM.

The answer would be B.

7 0

3 years ago

A cable company charges $70 per month for cable service. Please help ASAP Check screenshot pleasee (20 points) Thank youu! :)) &

LiRa [457]

Answer:

for #1: 70x = y

for #2: 70x + 100 = y

Step-by-step explanation:

you pay $70 for x amount of time with will equal your total therefore it's 70x = y

if you have to pay a one-time fee of $100.00 then you have to add that to your equation giving you 70x + 100 = y

hope this is right

hope it helps too :)

8 0

3 years ago

What is the radius of a hemisphere with a volume of 6466 to the nearest tenth of a meter?

Alecsey [184]

Answer:

14.6m

Step-by-step explanation:

volume of a hemisphere = (2/3)πr3

Therefore r = Cube root of ( Volume * 3/2 * 1/ pi)

r = cube root ( 6466 * 3/2 * 1/ π)

r = 14.56

in the nearest tenth of a meter = 14.6 m

7 0

3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

3 years ago