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morpeh [17]
3 years ago
10

Ms.green earns $908 per week for 40 hours of work. She earns 1.5 times her hourly rate for working any hours over 40 hours in a

week. How much money did Ms.green earn this week for working 50 hours? Show your work
Mathematics
2 answers:
Maurinko [17]3 years ago
6 0
908÷40=22.70

22.70÷2=11.35

11.35+22.70=34.05

34.05*10=340.50

908+340.50=1,248.50

1,248.50 for 50 hours
RSB [31]3 years ago
3 0
Divide 908 by 40 to find her hourly rate: 22.70
1.5 times 22.70: 34.05 (this is what she gets paid for each hour over 40)

50 hours = 40 hours + 10 hours (overtime)
                   908 + 10(34.05)
               =  1248.50
The answer is $1248.50
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Determine whether PQ and UV are parallel,
lys-0071 [83]

Answer:

The answer is below

Step-by-step explanation:

The slope of a line (m) is given by:

m=\frac{y_2-y_1}{x_2-x_1}

Two lines are parallel if they have the same slope and perpendicular if the product of their slope is -1.

1)

Slope\ of\ PQ=\frac{1-(-2)}{9-(-3)}=\frac{1}{4}  \\\\Slope\ of\ UV=\frac{-2-6}{5-3}=-4

Since the product of their slope is -1, they are perpendicular

2)

Slope\ of\ PQ=\frac{1-7}{2-(-10)}=-\frac{1}{2}  \\\\Slope\ of\ UV=\frac{1-0}{6-4}=\frac{1}{2}

Since the slope is not the same or product of their slope is not -1, they are neither parallel or perpendicular

3)

Slope\ of\ PQ=\frac{8-1}{9-1}=\frac{7}{8}  \\\\Slope\ of\ UV=\frac{8-1}{2-(-6)}=\frac{7}{8}

Since the slopes are the same, they are parallel

4)

Slope\ of\ PQ=\frac{3-0}{9-(-4)}=\frac{3}{4}  \\\\Slope\ of\ UV=\frac{6-(-3)}{8-(-4)}=\frac{3}{4}

Since the slopes are the same, they are parallel

5)

Slope\ of\ PQ=\frac{1-2}{0-(-9)}=-\frac{1}{9}  \\\\Slope\ of\ UV=\frac{-1-8}{-2-(-1)}=9

Since the product of their slope is -1, they are perpendicular

6 0
3 years ago
Read 2 more answers
1-cos2a/1+cosa if 1/3 how much sin²a
Helen [10]

Answer:

(1-cos2A) /(1+cos2A) =tan²A

Proof:

We know that,

cos(A+B) =cosA.cosB-sinA.sinB

=>cos2A=cos(A+A)

=>cos2A=cosA.cosA - sinA.sinA

=>cos2A=cos²A-sin²A

=>cos2A=(cos²A-sin²A)/(cos²A+sin²A

Since {cos²A+sin²A=1}

Divide the numerator & the denominator by (cos²A) to get,

cos2A = {(cos²A-sin²A) ÷cos²A} / {(cos²A+sin²A) ÷cos²A}

cos2A ={(1-tan²A)/(1+tan²A)}

Then,

1-cos2A = 1-[{(1–tan²A)/(1+tan²A)}]

1-cos2A =(1+tan²A-1+tan²A)/(1+tan²A)

1-cos2A=(2tan²A)/(1+tan²A)

And now.......

1+cos2A=1+[{(1-tan²A)/(1+tan²A)}]

1+cos2A={1+tan²A+1-tan²A}/{1+tan²A}

1+cos2A=2/(1+tan²A)

So now,

(1-cos2A)/(1+cos2A)= {2tan²A/(1+tan²A)}÷{2/(1+tan²A)}

={(2tan²A)(1+tan²A)}÷{2(1+tan²A)}

=tan²A

Step-by-step explanation:

make me as brain liest

3 0
4 years ago
8 – 2x = 20<br> What is x
Nesterboy [21]

Here,

8-2x=20

or,8-20=2x

or, -12=2x

or, x=-12/2

therefore,

x=-6

8 0
3 years ago
A large candy bar was 147 calories. The smaller candy bar has 29 percent less
harina [27]
104.37 , because if you do 147•29% it is equivalent to 42.63 and then if you do 147-42.63=104.37
5 0
3 years ago
Write an expression for angle rst.
levacccp [35]

Answer:

∠RST = 6x - 18

Step-by-step explanation:

Since SQ bisects ∠TSR then ∠RSQ = ∠QST = 3x - 9

and ∠RST = ∠RSQ + ∠QST = 3x - 9 + 3x - 9 = 6x - 18


4 0
4 years ago
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