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2 years ago

8

Solve for x: one over eight (8x + 15) = 24

2 answers:

Dahasolnce [82]2 years ago

5 0

Answer:

177/8

Step-by-step explanation:

You're welcome beautiful!<3

svetlana [45]2 years ago

4 0

1/8 ( 8x + 15)= 24

x + 1 7/8= 24

x= 24 - 1 7/8

x= 22 1/8

The answer is none of them.

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A geometry teacher asked Radric to define "parallel lines." Radric said that parallel lines are lines in a plane that do not hav

UkoKoshka [18]

Answer: The answer is Yes.

Step-by-step explanation: Given in the question that Radric was asked to define "parallel lines" and he said that parallel lines are lines in a plane that do not have any points in common. We are to decide whether Radric's definition is valid or not.

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2 years ago

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Which relationships have the same constant of proportionality between y and x as in the equation y=1/3x?

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Judy’s brother Sam has a collection of 96, but what are the 10 way Sam can divide his comic books equal groups

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1 year ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

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Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

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4 0

2 years ago

How long will it take before twenty percent of our 1,000-gram sample of uranium-235 has decayed? [See Section 6.6 Example 13] Th

horsena [70]

Answer:

From what I can see of the problem, you CANNOT solve for the half-life of U 235 AND then solve for the length of time to determine the 20% decay.

U-235 half-life is 704,000,000 years. (Wikipedia)

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elapsed time = 7.04 x 10^8 * [log (100 % / 80%) / log 2]

elapsed time =7.04 x 10^8 * [log (1.25) / .30103]

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elapsed time = 7.04 x 10^8 * 0.321928047

elapsed time = 226,637,000 years

Step-by-step explanation:

8 0

1 year ago