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2 years ago

Question

Mathematics

1 answer:

damaskus [11]2 years ago

3 0

Answer:

10 $20 dollar bills, and 2 $5 dollar bills

Really, you could say that he has 9 twenties, and 4 fives, but that would be impractical.

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Please can someone help me with this question

sasho [114]

Answer:

r = - $\frac{6a}{1-5m^2}$

Step-by-step explanation:

Rewrite the equation as $\sqrt{x} 6a+r/5r$ = m

Remove the radical on the left side of the equation by squaring both sides of the equation.

( $\sqrt{\frac{6a+r}{5r} )^{2}$ = m^2

Then, you simplify each of the equation.

Rewrite: ( $\sqrt{\frac{6a+r}{5r} )^{2}$ as $\frac{6a+r}{5r} = m^{2}$

Remove any parentheses if needed.

Solve for r.

Multiply each term by r and simplify."

Multiply both sides of the equation by 5.

6a+r= m^2r⋅(5)

Remove parentheses.

Move 5 to the left of (m ^2) r

6a+r=5m^2)r

Subtract 5m^2)r from both sides of the equation.

6a+r-5m^2)r=0

Subtract 6a from both sides of the equation.

r-5m^2)r=-6a

Factor r out of r-5m^2)r

r(1-5m^2)=-6a

Divide each term by 1-5m^2 and simplify.

r = - $\frac{6a}{1-5m^2}$

There you go, hope this helps!

4 0

3 years ago

Ms. Lou rolls out 2 3/5 ft. of dough. If she cuts the dough into sections that are 1/4 of a ft. each, how many sections of dough

Natasha_Volkova [10]

Answer:

Step-by-step explanation:

We need to divide 2 3/5 ft by 1/4 ft.

2 3/5 / 1/4 =

= 2 3/5 * 4

= 13/5 * 4

= 52/5

= 10 4/10

Answer: 10

5 0

2 years ago

What is the perimeter of a triangle with sides the following lengths: 3a + 2c,5c + 9 and 5a - 6. Simplify the expression.

Svetach [21]

Answer:

8a+7c-3

Step-by-step explanation:

8 0

2 years ago

What eigen value for this matix (1 -2) (-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

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

5 0

2 years ago

Read 2 more answers

5x-3x+7=3x-1 <img src="https://tex.z-dn.net/?f=4r%20%2B%20%20%5Cfrac%7B2%7D%7B4%7D%20r%20%2B%2014%20%3D%205r%20-%20%20%5Cfrac

nirvana33 [79]

1). 5x-3x+7 = 3x-1

2x + 7 = 3x - 1

3x-2x=7+1

x=8

$4r + \frac{2}{4} r + 14 = 5r - \frac{3}{4} r + 1 - 3\\ \\5r-4r-\frac{3}{4} r - \frac{2}{4} r = 14-1+3\\ \\r-\frac{5}{4}r =16\\ \\\frac{4}{4}r-\frac{5}{4}r =16\\ \\-\frac{1}{4}r=16\\ \\ r=-64$

5 0

3 years ago