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Eva8 [605]
1 year ago
8

helene was paid a total of $33.00 for the 6 hours she worked. she continues to be paid at the same rate. how long would it take

her to earn $165.00 for working?
Mathematics
1 answer:
natita [175]1 year ago
7 0

The number of hours that helena need to work to generate 165 dollars is

33 hours.

Helena has to work hours to earn $165. 33

Algebraic equations are equations with

unknown variables. The alphabet letters are used to represent algebraic equations.

Total $33.00 in her 6 hours she worked.

This question should find out how long it takes her to earn her $165.00 in that job.

let x hours be required work hours to earn $165.00 ..

This can be expressed in algebraic equations as:

x= 6× 165/33

x= 990/33

x= 30

Hence, helna have to work thirty hours to earn $165.00 ..

To learn more about Work time problem solving, refer:

brainly.com/question/15447610

#SPJ4

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I need help with these 2 problems I cant figure out.
Alla [95]

I started by labeling the right angle (Angle C) 90º. Next, I wrote down everything in one equation.

2x + 90 + 3x - 20 = 180º (180 degrees in a triangle)

Next, I add 20 on both sides.

2x + 90 + 3x = 200º

I combine like terms (2x and 3x)

5x + 90 = 200º

I subtract 90 from both sides.

5x = 110º

Divide 110 by 5 to get x.

x = 22

======

For problem two, I label all the angles I know.

49º + 80º + r = 180º

I add 80 and 49.

129º + r = 180º

I subtract 180 and 129 and get 51º, which is your angle for R.

For angle X, you know that angle R plus angle X equals half of a circle, which is 180º

We know from before that 129º is 180º without R, so X is 129º

I hope this helps! Let me know if I'm wrong!

5 0
2 years ago
The math club members were throwing a pizza party and each person had the choice of two slices one topping. 1/2 of the students
Andrew [12]
1/8 = 4 so then just multiply by 8 to get 32
8 0
2 years ago
Read 2 more answers
What is a non-example of a independent variable?
alexandr402 [8]
An non-example of a independent variable is how much money you make selling cookies , because it depends on the number of cookies you sell .
7 0
3 years ago
Read 2 more answers
There is a competition in the cinema to win free tickets. You must guess the ages of the four employees. Here are the clues: - W
jenyasd209 [6]

Answer:

Kirk is <u>28</u> years old, Brian is <u>36</u> years old, Matt is <u>18</u> years old, and the manager is <u>24</u> years old.

Step-by-step explanation:

Let k be Kirk's age.

Let b be Brian's age.

Let m be Matt's age.

Let m_{1} be the boss/manager's age.

From the information given, we can set up 4 equations:

k+b+m+m_{1}=106

k=2(m_{1}-10)

b=2m_{1}-12

m=\frac{1}{2}m_{1}+6

Rewrite the first equation, substituting k, b, and m in terms of m_{1} to get:

2(m_{1}-10)+(2m_{1}-12)+(\frac{1}{2}m_{1}+6)+m_{1}=106

Open up the parentheses using the distributive property (which is a(b-c)=ab-ac) and combine like terms to get:

5.5m_{1}-26=106

Add 26 to both sides to reach:

5.5m_{1}=132

Thus, m_{1}=24. Substitute 24 for m_{1} in the second, third, and fourth original equations to find that k=28, b=36, and m=18. Therefore, Kirk is <u>28</u> years old, Brian is <u>36</u> years old, Matt is <u>18</u> years old, and the manager is <u>24</u> years old. To check, you can add up all of the ages to get 106.

Hope this helps :)

8 0
2 years ago
Algebra public opinion polls reported in newspapers are usually given with a margin of error. for example, a poll with a margin
sweet-ann [11.9K]

Answer:

Minimum percent of the vote that candidate Towne is expected to recieve:

m=51% - 6.3% * 51% =47.787%

Maximum percent of the vote that candidate Towne is expected to recieve:

M=51% + 6.3% * 51% = 54.213%

Solution:

Margin of error: E=6.3%

Minimum percent of the vote that candidate Towne is expected to recieve:

m=51% - E * 51%

m=51% - 6.3% * 51%

m=51% - 51% * 6.3 / 100

m=51% - 3.213%

m=47.787%

Maximum percent of the vote that candidate Towne is expected to recieve:

M=51% + E * 51%

M=51% + 6.3% * 51%

M=51% + 51% * 6.3 / 100

M=51% + 3.213%

M=54.213%


8 0
3 years ago
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