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2 years ago

PLEASE HELP ASAP!! THIS IS LATE! :(

Mathematics

1 answer:

Svetach [21]2 years ago

4 0

Answer:

States sent representatives to Congress and provided soldiers and some officers to protect the country.

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Snowcat [4.5K]

Answer: y= -14/27

Explanation: Calculator

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3 years ago

Help ASAP 2k-6-7+3k=-48

k0ka [10]

Answer:

k=-7

Step-by-step explanation:

2k-6-7+3k=-48

Combine like terms

5k -13 = -48

Add 13 to each side

5k-13+13 = -48+13

5k = -35

Divide by 5

5k/5 = -35/5

k =-7

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3 years ago

Read 2 more answers

The probability that Carmen will get an A on her math test is 80%, and the probability that she will get an A on her science tes

VLD [36.1K]

Answer:

70%

Step-by-step explanation:

60+80=140

140/2=70

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3 years 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:

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

$8r = \ln{1.79}$

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

$r = 0.0726$

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).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

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

4 0

3 years ago

Simplify<br> x^2-9/x^2+x-12

vagabundo [1.1K]

Answer:

I love algebra anyways

the ans is in the picture with the steps

(hope it helps can i plz have brainlist :D hehe)

Step-by-step explanation:

3 0

3 years ago