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2 years ago

If mark has 3 friends help him paint his garage it will take 60 minutes how long will it take if he has 4 friends help him

Mathematics

1 answer:

olga2289 [7]2 years ago

4 0

Answer:

40 minutes

Step-by-step explanation:

hope this will help :)

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Kim spent $33 on a magazine and

loris [4]

Answer:

Seven candy bars.

Explanation:

Suppose the number of candy bars purchased is x

The questions say he bought A magazine, means one magazine

the equation is:

4x + 5 = 33

4x = 33 – 5

4x = 28

x = 7

Kim bought 7 candy bars.

Hope this helps! :)

5 0

2 years ago

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The graph shows the cost per day of electricity over a seven-day period.

Darina [25.2K]

A. Negative. You’re welcome

5 0

2 years ago

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The length of a rectangle is 4 more than twice the width. The perimeter is 74 feet. Find the length and width of the rectangle.

mafiozo [28]

(2*length)+(2*width)=perimeter
length=4+2w
(2(4+2w))+(2(w))=74
74=(8+4w)+(2w)
74=6w+8
66=6w
w=11
length=4+2(11)
l=4+22
l=26

3 0

2 years ago

Graph y = -x^2 - 2. Identify the vertex of the graph. Tell whether It is a minimum a<br> or maximum

lawyer [7]

Answer:

(0,-2)

maximum

Step-by-step explanation:

-x∧2-2 = -1 (x∧2) -2

so the vertex should be (0,-2)

As it is negative for the x∧2, so the graph opens downwards. So, the vertex is a maximum.

5 0

2 years ago

What eigen value for this matix <br> (1 -2)<br> (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$

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2 years ago

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