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4 years ago

7

kevin had 13 coins in his pocket. all of his coins are either dimes or quarters. when he emptied his pocket, kevin found that he

had 2.05. how many dimes and quarters does he have?

2 answers:

Gnesinka [82]4 years ago

7 0

5 quarters and 8 dimes = $2.05

WINSTONCH [101]4 years ago

5 0

Answer:

Kevin has 5 quarters and 8 dimes.

Step-by-step explanation:

First of all, a quarter is worth 0.25 dollars, while a dime is worth 0.10 dollars. As Kevin has 13 coins in his pocket which in sum make $2.05, and all of them are either dimes or quarters, in order to determine how many dimes and quarters does he have, we have to take into account that he's got an odd number of quarters, since the final amount of money sums $2.05, that is, it ends in 5.

Therefore, if he had 1 quarter and 12 dimes (0.25 + 12x0.10 = 1.45), he would have $1.45.

If he had 3 quarters and 10 dimes (0.25x3 + 10x0.10 = 1.75), he would have $1.75.

If he had 5 quarters ahd 8 dimes (0.25x5 + 8x0.10 = 2.05), he would have $2.05, so this is the correct option.

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Step-by-step explanation:

Given data:

Annual dividend per share = $ 2.30

Required return on the preferred stock = 6.5 %

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on substituting the respective values, we get

Selling price of the preferred stock = $ 2.30 / 6.5% = $ 2.30 / 0.065

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3 years ago

Use properties to rewrite the given equation. Which equations have the same solution as 3/5x +2/3 + x = 1/2– 1/5x? Check all tha

vodomira [7]

we have

$\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x$

Combine like terms in both sides

$(\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

we know that

$(\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x$

substitute in the expression above

$\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$ -----> equation A

Multiply equation A by $5*3*2=30$ both sides

$30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)$

$48x+20=15-6x$ ---------> equation B

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$48x+6x=15-20$

$54x=-5$ ---------> equation C

Solve for x

$x=-\frac{5}{54} =-0.09$

We are going to proceed to verify each case to determine the solution.

<u>Case a)</u> $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

the case a) is equal to the equation A

so

the case a) have the same solution that the given equation

<u>Case b)</u> $18x+20+30x=15-6x$

Combine like terms in left side

$(18x+30x)+20=15-6x$

$(48x)+20=15-6x$

the case b) is equal to the equation B

so

the case b) have the same solution that the given equation

<u>Case c)</u> $18x+20+x=15-6x$

Combine like terms in left side

$(18x+x)+20=15-6x$

$(19x)+20=15-6x$

$19x+6x=15-20\\25x=-5\\x=-0.20$

$-0.20\neq -0.09$

therefore

the case c) not have the same solution that the given equation

<u>Case d)</u> $24x+30x=-5$

Combine like terms in left side

$54x=-5$

the case d) is equal to the equation C

so

the case d) have the same solution that the given equation

<u>Case e)</u> $12x+30x=-5$

Combine like terms in left side

$42x=-5$

$x=-5/42=-0.12$

$-0.12\neq -0.09$

therefore

the case e) not have the same solution that the given equation

therefore

<u>the answer is</u>

case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

case b) $18x+20+30x=15-6x$

case d) $24x+30x=-5$

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