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4 years ago

5

Kay and Allen sell cell phones. Kay sold 17 more cell phones than Allen. Together they sold 117. How many did they sell individu

ally. Show the equation

1 answer:

denis-greek [22]4 years ago

5 0

Answer:

Kay sold 67 cell phones and Allen sold 50 cell phones.

Step-by-step explanation:

Let k = # cell phones Kay sold and a = # cell phones Allen sold.

k + a = 117

k = 17 + a

So:

(17 + a) + a = 117

2a = 100

a = 50

Plug this back into 2nd equation:

k = 17 + a

k = 17 + 50

k = 67

Kay sold 67 cell phones and Allen sold 50 cell phones.

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3 years ago

Ms. Garza invested $50,000 in three different accounts. She invested three times as

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Answer:

The amount invested in the account that paid 8% was $18,000

The amount invested in the account that paid 10% was $6,000

The amount invested in the account that paid 12% was $26,000

Step-by-step explanation:

Let

3x -----> the amount invested in the account that paid 8%

x -----> the amount invested in the account that paid 10%

$50,000-4x ----> the amount invested in the account that paid 12%

in this problem we have

$8\%=8/100=0.08\\10\%=10/100=0.10\\12\%=12/100=0.12$

we know that

$3x(0.08)+x(0.10)+(50,000-4x)(0.12)=5,160$

Solve for x

$0.24x+0.10x+6,000-0.48x=5,160$

$0.48x-0.24x-0.10x=6,000-5,160$

$0.14x=840$

$x=\$6,000$

$3x=\$18,000$

$\$50,000-4x=\$26,000$

therefore

The amount invested in the account that paid 8% was $18,000

The amount invested in the account that paid 10% was $6,000

The amount invested in the account that paid 12% was $26,000

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3 years ago

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Answer:

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Step-by-step explanation:

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I got this answer with complete explanation from Gauthmath app. They have the tutors who can answer our questions. In my opinion it's a very good app to get help. You don't even need to wait. I think you should check it out once.

<em>Hope </em><em>my </em><em>answer </em><em>helps.</em><em> </em><em>Best </em><em>of </em><em>luck!</em><em> </em><em />

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Solnce55 [7]

Answer :

(1) $\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) $\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}$

(3) $\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}$

(4) $\frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}$

(5) $\frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}$

(6) $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}$

Step-by-step explanation :

(1) The given expression is: $\frac{1}{x}+\frac{1}{y}$

$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) The given expression is: $\frac{1}{5x}+\frac{1}{3y}$

$\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}$

(3) The given expression is: $\frac{1}{7x}-\frac{1}{2y}$

$\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}$

(4) The given expression is: $\frac{2}{5x}+\frac{3}{7y}$

$\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}$

(5) The given expression is: $\frac{7}{11x}-\frac{1}{33y}$

$\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}$

(6) The given expression is: $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}$

$\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}$

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3 years ago