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3 years ago

8

Fill in the blank distributive property

1 answer:

Sav [38]3 years ago

3 0

$4(20+1)=4(20)+4(1)=80+4=84$

It's still just easier to add first though.

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What is the middle of 1 and 6

devlian [24]

Answer:

The middle between two numbers is the average of those numbers. So we add those two numbers together and divide by two. And the answer is (1 + 6) / 2 = 3.5

5 0

2 years ago

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PLEASE HELP MEEEE HURRRY!!! :)

sammy [17]

Answer:

Option D

Step-by-step explanation:

We are given the following equations -

$\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}$

It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -

$\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}$ First, I swapped the first and third rows.

$\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}$ Leading coefficient of row 2 canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}$ The start value of row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}$ Matrix rows 2 and 3 were swapped.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}$ Leading coefficient in row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}$

And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -

$\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}$

The positioning of the zeros indicated that there was no solution!

<u><em>Hope that helps!</em></u>

6 0

3 years ago

A lone segment has ______ endpoints.<br><br>1) three<br>2)two<br>3)zero<br>4)one

Nikolay [14]

Answer:

<h2><em><u>2)</u></em><em><u> </u></em><em><u>two</u></em></h2>

Step-by-step explanation:

A line segment forms when it has two end points

8 0

3 years ago

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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

Random groups of 30 teachers were asked the starting annual salary for their

Ilya [14]

Answer:

B

Step-by-step explanation:

please mark as brainlist answer

5 0

3 years ago

Read 2 more answers