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vampirchik [111]
2 years ago
14

HOW AM I SUPPOSED TO DO THIS IF YOU DONT KNOW HOW TO DO THIS DONT HELP AN GET ME SOMEONE THAT WILL someone help me

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
6 0
Just take each number in the first table and divide it by the total then x by 100

Eg the first one 24/120 x 100 = 20%
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One month, ruby worked 6 more hours than Issac, and Svetlana worked 4 times as many hours as Ruby. Together they worked 126 hour
garik1379 [7]
Ruby's hours will represent x+6
Svetlana's hours will represent 4(x+6)
Issac will represent x.

The equation would read:
x+(x+6)+(4(x+6))=126
x+x+6+4x+24=126
6x+30=126
Subtract 30 from both sides.
6x=96
Divide 6 on both sides to get x by itself.
x=16

So:
Issac worked 16 hours
Ruby worked 22 hours
Svetlana worked 88 hours.

To check your answer, plug in the values:
16+22+88=126
126=126

Hope this helps.
8 0
3 years ago
Determine if there is a proportional relationship between the length of the diagonals and the perimeters of the squares.
garri49 [273]

Answer:

that polygon. Since the ratio P/s is constant, hence we can say that the perimeter of a square is directly proportional to the length of a side. Hence the constant of proportionality for the perimeter of a square and length of it's side is 4.08

Step-by-step explanation:

where's the question and picture

5 0
2 years ago
Read 2 more answers
The area of a triangle is 150 cm^2. If the area of a triangle is given by the formula A=0/5b*h, write an equation which relates
taurus [48]

Answer:

Step-by-step explanation:

If the area of a triangle is 150cm^2, the formula would look like:

150=0.5b*h

So, if we were to solve for the base in terms of h, we'd simply divide both sides by 0.5b getting:

(150)/(0.5b)=h

which simplifies to

h=300b

4 0
3 years ago
I need help on these two! Please and thank you! (:
ZanzabumX [31]

Answer:

Answers and methord in the pic

5 0
2 years ago
Read 2 more answers
Simplify.<br> 4/x^2-35 - 1/x+5
Oksana_A [137]
If I understand your question well, I suppose you mean \frac{4}{ x^{2} -35} -  \frac{1}{x+5}
\frac{4}{ x^{2} -35} - \frac{1}{x+5} =  \frac{4}{(x+5)(x-7)}- \frac{1}{x+5} \\ = \frac{1}{x+5}( \frac{4}{x-7}-1) \\ = \frac{1}{x+5}( \frac{4}{x-7}- \frac{x-7}{x-7}) \\ = \frac{1}{x+5}( \frac{4-x+7}{x-7}) \\ = \frac{1}{x+5}( \frac{11-x}{x-7}) \\ = \frac{11-x}{ x^{2} -35}
6 0
3 years ago
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