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lyudmila [28]
3 years ago
9

What would I put to show my work?

Mathematics
1 answer:
fredd [130]3 years ago
6 0

Answer:

x = 23

Step-by-step explanation:

These are supplementary angles so the sum of those angles add up to 180:

3x + 16 + 5x - 20 = 180

8x - 4 = 180

8x = 180 + 4

8x = 184

x = 184/8

x = 23

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666 + 667 + 504 times 500 divided by 5 = ?
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one positive number is 28 less than another number. if the reciprocal of the smaller number is added to four times the reciproca
marishachu [46]
Let x represent the larger number.
  1/(x -28) +4/x = 1/6
Multiplying by the product of denominators, we have
  6x +24(x -28) = x(x -28)
In standard form, this is
  x^2 -58x +672 = 0
  (x -42)(x -16) = 0

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7 0
3 years ago
Right the point-slope form of the line that passes through negative (7,8) and has a slope of -5
Delicious77 [7]

Answer:

y-8=-5(x-7)

Step-by-step explanation:

we know that

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y-y1=m(x-x1)

we have

m=-5\\(x1,y1)=(7,8)

substitute

y-8=-5(x-7)

4 0
4 years ago
Solve the initial value problem
tigry1 [53]

9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}

Look for an integrating factor \mu(t):

\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}

Multiply both sides by \mu:

(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}

Condense the left side as the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}

Integrate both sides:

(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt

For the integral on the right, substitute

u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du

\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du

\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C

\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)

\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C

\implies y=7(t+1)+2+C(t+1)^{7/9}=7t+9+C(t+1)^{7/9}

Given that y(0)=12, we get

12=9+C\implies C=3

\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}

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Y and X is urgent answer
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