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aniked [119]
3 years ago
12

To paint a house, a painting company charges a flat rate of $500 for supplies plus $50 for each hour of labor. What is the slope

of the line?

Mathematics
1 answer:
Nataly_w [17]3 years ago
5 0
The slope of the line is 50. You can try looking at the equation, which is

y=50x+500

500 is your y-intercept, and is where the cost starts.

50 is your slope, since the cost goes up by $50 every time.

x is how many hours, which determines how many $50s there are.

I hope this helps. All in all, the slope is 50.
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Please Help with this Problem!
tatiyna

Answer: 3.5, 4.5, 9.5, 3.5

Step-by-step explanation:

Look at the image below to see where A, B, C, and D are.

A + B = 8

B + D = 8

A + C = 13

C - D = 6

we can see that A + B = 8 and D + B = 8,  so A = D

substitute this into A + C = 13 to get D + C = 13

from D + C = 13 we can get D = 13 - C

plug this into C - D = 6 to get C - (13 - C) = 6

2C - 13 = 6

2C = 19

C = 9.5

Now we can find D = 13 - C = 13 - 9.5 = 3.5

D = 3.5

Now we can find A = D = 3.5

A = 3.5

Now we can find B from A + B = 8

B = 8 - A = 8 - 4.5 = 4.5

B = 4.5

6 0
3 years ago
What the square root of 4
Aliun [14]

Answer:

2

Step-by-step explanation:

2x2=4

3 0
4 years ago
Read 2 more answers
Let the equation C= 2.32 N + 34,180 represent the cost of raising a child, C on an income, N. If the Corlone family has an incom
Genrish500 [490]

Answer:

$126,980

Step-by-step explanation:

Given the equation C = 2.32N + 34,180 where 'C' is the cost of raising a child and 'N' is the income.  If the Corlone family has an income of $40,000, you can use this value for 'N' and solve for 'C':

C = 2.32(40,000) + 34,180

C = 92,800 + 34,180

C = $126,980

8 0
4 years ago
Can someone help with this pleasee?
enot [183]

Answer:

That is so easyyyyyyyyy

Step-by-step explanation:

3 0
3 years ago
) a, p and d are n×n matrices. check the true statements below:
BigorU [14]
A. False. Consider the identity matrix, which is diagonalizable (it's already diagonal) but all its eigenvalues are the same (1).

b. True. Suppose \mathbf P is the matrix of the eigenvectors of \mathbf A, and \mathbf D is the diagonal matrix of the eigenvalues of \mathbf A:


\mathbf P=\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{bmatrix}

\mathbf D=\begin{bmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{bmatrix}

Then

\mathbf{AP}=\begin{bmatrix}\mathbf{Av}_1&\cdots&\mathbf{Av}_n\end{bmatrix}=\begin{bmatrix}\lambda_1\mathbf v_1&\cdots&\lambda_n\mathbf v_n\end{bmatrix}=\mathbf{PD}

In other words, the columns of \mathbf{AP} are \mathbf{Av}_i, which are identically \lambda_i\mathbf v_i, and these are the columns of \mathbf{PD}.

c. False. A counterexample is the matrix

\begin{bmatrix}1&1\\0&1\end{bmatrix}

which is nonsingular, but it has only one eigenvalue.

d. False. Consider the matrix

\begin{bmatrix}0&1\\0&0\end{bmatrix}

with eigenvalue \lambda=0 and eigenvector \begin{bmatrix}k&0\end{bmatrix}^\top, where k\in\mathbb R. But the matrix can't be diagonalized.
7 0
3 years ago
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