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Veronika [31]
2 years ago
15

There are 130 m&m's in a bag. there are half as many green as red. the number of blue is equal to the number of red and gree

n combined. there are half as many yellow as blue. there are 4 more brown than yellow. how many of each color are in the bag? what is the answer
Mathematics
1 answer:
AleksandrR [38]2 years ago
8 0
Let x = red
Let 0.5x = green
Let (x) + (0.5x) = 1.5x = blue
Let 0.5(1.5x) = 0.75x = yellow
Let 0.75x + 4 = brown

(x) + (0.5x) + (1.5x) + (0.75x) + (0.75 + 4) = 130
x + 0.5x+ 1.5x + 0.75x + 0.75 + 4 = 130
4.5x + 4 = 130
4.5x = 126
x = 28

Use this value to calculate the other required numbers.
red = 28
green = 0.5x = 0.5(28) = 14
blue = 1.5x = 1.5(28) = 42
yellow = 0.75x = 0.75(28) = 21
brown = 0.75x + 4 = 0.75(28) + 4 = 25
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No, you would go into the negatives. Eventually, you would own him more at night than you got in the morning. Don't work for the devil.

Step-by-step explanation:

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3 years ago
70 POINTSSSS HELPPPP !
Radda [10]
40 is the answer for your question
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2 years ago
Please help me solve this... I will mark you brainliest if it’s correct
Crank
(5 + 7x + 6)/2 = 3x + 9

7x + 11 = 6x + 18

7x - 6x = 7

x = 7

EF = 3x + 9

= 3(7) + 9

= 21 + 9

= 30
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Pleases someone explain
Nikolay [14]

When you have a negative exponent, you move the base with the negative exponent to the other side of the fraction to make the exponent positive.

For example:

\frac{1}{y^{-2}} =\frac{y^2}{1} or y^2

2x^{-3} or \frac{2^1x^{-3}}{1} =\frac{2}{x^3}

\frac{1}{x^{-5}} =x^5

When a base with an exponent is divided by a base with an exponent, you subtract the exponents together. (But you can only combine the exponents when the bases are the same)

For example:

\frac{x^2}{y}   (can't combine because they have different bases of y and x)

\frac{x^5}{x^3} =x^{(5-3)}=x^2

\frac{2^2}{2} =2^{(2-1)}=2^1 = 2

When you multiply an exponent directly to a base with an exponent, you multiply the exponents together.

For example:

(x^2)^3=x^{(2*3)}=x^6

(y^2)^{10}=y^{(2*10)}=y^{20}

\frac{2e^0}{(e^{-3})^2}     First multiply the exponents together in the denominator

\frac{2e^0}{e^{(-3*2)}}= \frac{2e^0}{e^{-6}}        Now subtract the exponents together

2e^{(0-(-6))}  (two negative signs cancel each other out and become positive)

2e^{(0+6)}

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AlladinOne [14]
N=4 have a good day



:))
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