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Irina-Kira [14]

2 years ago

15

Find the greatest common factor for the terms 6x^5y^8, 8x^3y^9, 12x^2y^4

1 answer:

Alla [95]2 years ago

4 0

Answer:

i dont know

Step-by-step explanation: just for points sorry :(

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I need help in solving this, I dont want an answer. 10. The Wilson family had a family portrait framed for their living room. Th

ollegr [7]

So to find an area of a square is length times width. 45 times 37 =1665inches now convert that to ft and that’s your answer!
Hope this helps!

3 0

3 years ago

What is the determinant of k =[ 6 8]

Sergeeva-Olga [200]

That would be 6 *3 - 8*0 = 18

Its D

5 0

3 years ago

Given: ABCD is a ∥-gram; DE ∩ AB =F <br> Prove: △ADF∼△CDE

Vedmedyk [2.9K]

Answer:

By AA similarity

Step-by-step explanation:

We have been given that ABCD is a parallelogram

So, by the property of parallelogram AB ||CD and FD is cutting the line BC

Hence, FD is transverse line. In transverse line alternate angles are equal.

Therefore, ∠AFD=∠EDC (alternate interior angles)

And ∠FAD=∠ECD (opposite angles in parallelogram)

Therefore, by AA similarity △ADF∼△CDE

5 0

2 years ago

Find the x-intercept of the rational function.

alexira [117]

Answer:

What question are we supposed to answer

Step-by-step explanation:

Sorry I read the whole thing multiple times and couldn’t find what the question was

If you wanted the points to be graphed use Desmo calculator they graph the points for you:)

8 0

2 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago