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3 years ago

9

Can y’all help me on this math question?!

2 answers:

Burka [1]3 years ago

6 0

Answ you tell me

Step-by-syou tell me

SIZIF [17.4K]3 years ago

4 0

Line segment VU= 34, choice A

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If ∠A and ∠B are supplementary and m∠A = 37°45', then m∠B = ______.

xenn [34]

Answer: 142°15'

Explanation:

Supplementary angles are angles that add up to make a straight angle ( $180^{\circ}$ ).

Therefore, we must find the angle ∠B such that

∠A + ∠B = 180° (1)

We know that ∠A = 37°45', so we can re-arrange equation (1) to find the magnitude of ∠B:

∠B = 180° - ∠A = 180° - 37°45' = 142°15'

So, the correct answer is

142°15'

6 0

4 years ago

Read 2 more answers

Please help me fast x over 7 minus 12 equals negative 2

andrey2020 [161]

Answer:

x= 70

Step-by-step explanation:

The equation is: x/7-12= -2

First, do inverse operations. Do = +12 on each side to get rid of 12

Now: x/7=10

Multiply 7 on each side

Now: x=70

Hope this helped!!! :D

5 0

3 years ago

Read 2 more answers

Write an equation of the line passing through the points (4,6) and (-2, - 18).

olga_2 [115]

Answer:

y = 4x – 10

Step-by-step explanation:

slope formula: $m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

point-slope formula: $y-y_{1}=m\left(x-x_{1}\right)$

Use slope formula and plug in your numbers: $m = \frac{-18-6}{-2-4 }$
Solve for slope $m = 4$
Use the point-slope formula to solve for the equation: $y-6=4\left(x-4\right)$
Solve and find equation: $y=4x-10$

Note: When given two points you only need to pick one for the point-slope formula! Best to choose the positive one so you don't have to deal with the negatives!

Hope this helps! Please give Brainliest!

4 0

3 years ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago

Can you help me please

grigory [225]

m is 43

add my instagarm @jessica.letendre

5 0

4 years ago