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Oduvanchick [21]

3 years ago

6

Chris has twice as many $1 bills as

1 answer:

hodyreva [135]3 years ago

3 0

There is not enough information to give a numerical answer.
Let t=number of ten dollar bills.
"he has twice as many dimes as he has 10 dollar bills", so the amount A is
A=10*t+0.1*(2t)
"he has twice as many pennies as he has dimes"
A=10*t+0.1*(2t)+0.01*4t
"chris has twice as many 1 dollar bills as pennies"
A=10*(t)+0.1*(2t)+0.01*(4t)+1*(8t)
=18.24t
Therefore the total amount Chris has is 18.24 times the number of $10 bills he possesses.

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3 years ago

James bought the computer which was discounted from $350 to $273. What is the discount rate?

sasho [114]

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Which expression is equivalent to (81x4y16)12 for all positive values of x and y?<br> 40.5x4y16

IrinaVladis [17]

Answer:

$(81x^4y^{16})^{\frac{1}{2}} = 9 x^2 y^8$

Step-by-step explanation:

Given

$(81x^4y^{16})^{\frac{1}{2}}$

Required

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Express 81 as $9^2$

$(9^2x^4y^{16})^{\frac{1}{2}}$

Apply law of indices

$9^2^{\frac{1}{2}} * x^4^{\frac{1}{2}} * y^{16}^{\frac{1}{2}}$

$9^{\frac{2*1}{2}} * x^{\frac{4*1}{2}} * y^{\frac{16*1}{2}}$

$9^{\frac{2}{2}} * x^{\frac{4}{2}} * y^{\frac{16}{2}}$

$9^1 * x^2 * y^8$

$9 * x^2 * y^8$

$9 x^2 y^8$

Hence:

$(81x^4y^{16})^{\frac{1}{2}} = 9 x^2 y^8$

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2 years ago