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Nitella [24]
3 years ago
12

Lee wong is a sales salesperson at sears. she is paid 8.00 dollars per hour. plus a commission of 4%. assuming that she works 39

hours and has sales of 4,000, what is her gross pay?
Mathematics
1 answer:
velikii [3]3 years ago
6 0
39hours*8.00
312 dollars + 4,000sales*4%comission
312 dollars + 1600 dollars
1912 dollars
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You are selling tickets to a play. You have sold (3t+2) tickets for $5 each and (2t+5) tickets for $7 each. Write an expression
aliya0001 [1]

Given that the number of tickets that are sold for $5 is (3t+2) tickets.

The number of tickets that are sold for $7 is (2t+5) tickets.

We need to determine the expression that represents the total number of tickets sold and the simplified expression.

<u>Total number of tickets sold:</u>

The expression for (3t+2) tickets sold for $5 is given by

5(3t+2)

The expression for (2t+5) tickets sold for $7 is given by

7(2t+5)

The expression for the total number of tickets sold is given by

5(3t+2)+7(2t+5)

Thus, the expression that represents the total number of tickets sold is 5(3t+2)+7(2t+5)

<u>Simplified expression:</u>

We need to simplify the expression 5(3t+2)+7(2t+5)

Hence, let us multiply the terms within the bracket.

Thus, we get;

15t+10+14t+35

Adding the like terms, we have,

29t+45

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5 0
3 years ago
Find the quotient. The answer must be a proper fraction simplest form.<br> 4 13 ÷ −3 16
DerKrebs [107]

Answer:

5 or 5/1

Step-by-step explanation:

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3 years ago
What degree of rotation is represented on this matrix
Korvikt [17]

Answer:

Option B is correct

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Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Given the matrix: \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

Now, equate the given matrix with standard matrix we have;

\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} =  \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

On comparing we get;

\cos \theta = 0       and -\sin \theta =1  

As,we know:

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\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})

we get;

\theta = -90^{\circ}

and

\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})

we get;

\theta = -90^{\circ}

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Answer:

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That’s the answer to the question
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