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3 years ago

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1 answer:

UNO [17]3 years ago

7 0

What do you don’t get

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What is the value of m ? 2(3 m - 6 + 3 m) = 2 m + 2 -4 m

gulaghasi [49]

2(3m-6+3m)=2m+2-4m
6m-12+6m=2m+2-4m
12m-12=-2m+2
14m-12=2
14m=14
m=1

8 0

3 years ago

How do you solve 39 equal to 1 3/10b

elena-s [515]

If 39 = (1 3/10)b, we could simplify this by writing 39 - (13/10)b.

Solve this for b by mult. both terms by (10/13):

(10*39)
----------- = (10/13)(13/10)b. Then b = 30 (answer)
13

5 0

2 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.

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3 years ago

What is the value of the expression (2)^6?

Mumz [18]

The value of the expression (2)^6 is 64.

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3 years ago

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Which of the following options best describes the function graphed below?

wolverine [178]

Nonlinear increasing

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3 years ago