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Elena-2011 [213]
3 years ago
10

BRAINLIEST BRAINLIEST Please answer number 2 thank you

Mathematics
2 answers:
viktelen [127]3 years ago
7 0

Answer:

Wage = 7x,x ∈ [0,1,2,3,4]

Statements such as these can be converted into a mathematical language which is known as equations that can help us solve for the unknown. In this question, we need to find the wages for Charles who babysits. We have been given some statements which need to be translated into mathematical equations.

A typical equation has two sides, a right-hand side, and a left-hand side.

Let the total wage be given by  Wage  and the total number of hours worked to be given by  x  Thus, we can denote the function in functional notation as

Wage = 7x,x ∈ [0,1,2,3,4]

Wage = 7(4),4 ∈ [0,1,2,3,4] The symbol ∈ indicates set membership and means “is an element of”

Wage = 7(4),4 is an element/set of [0,1,2,3,4]

bija089 [108]3 years ago
4 0

Answer:

$28

Step-by-step explanation:

Time = 4 hours

Charge = $7 / hour

Multiply those two numbers:

4 hour × $7 / hour = $28

Function f(time) = $7 × time

Hopefully this answer helps you :)

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Find the area of the region defined by the region defined by the inequality 2|x| + 3|y-1| ≤ 6
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If x and y-1 have the same sign, then either

x>0,y>1 \implies 2|x| + 3|y-1| = 2x + 3(y-1)=6 \implies 2x + 3y = 9

or

x

If x and y-1 have opposite sign, then

x>0,y

or

x1 \implies 2|x| + 3|y-1| = -2x + 3(y-1) = 6 \implies 2x-3y = -9

This is to say that the region has boundaries given by these two sets of parallel lines, so we can equivalently describe the region with the set

R = \left\{(x,y) \mid -3\le2x+3y\le9 \text{ and } -9\le2x-3y\le3\right\}

The area of R is given by the double integral

\displaystyle \iint_R dx\,dy

To compute the area, change the variables to

\begin{cases}u = 2x + 3y \\ v = 2x - 3y\end{cases} \implies \begin{cases}x = \frac14(u+v) \\ y = \frac16(u-v)\end{cases}

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix}1/4 & 1/4 \\ 1/6 & -1/6\end{bmatrix}

with determinant \det(J) = -\frac1{12}. Then the integral transforms to

\displaystyle \iint_R dx\,dy = \iint_R |J| \, du \, dv = \frac1{12} \int_{-3}^9 \int_{-9}^3 dv\, du

which is 1/12 the area of a square with side length 12. Hence the integral evaluates to

\displaystyle \iint_R dx\,dy = \frac1{12}\times12^2 = \boxed{12}.

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Answer:

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Step-by-step explanation:

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9
barxatty [35]

Your equation for the first point will be y=x+50

Explanation...

The x is zero so we just only put x. The y is 50 so we input a 50 giving us y=x+50.

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Explanation...

Your slope is 2 so we input that and the y intercept is 100 giving us the equation y=2x+100.

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