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earnstyle [38]
3 years ago
14

Solve the equation. 2t +8= -10 solve for t​

Mathematics
2 answers:
Mazyrski [523]3 years ago
6 0

Answer:

Step-by-step explanation:

2t+8=-10, subtract 8 from each side

2t=-18, divide each side by 2

t=-9

Vesnalui [34]3 years ago
5 0

Answer:

t= -9

Step-by-step explanation:

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2(n - 1) + 4n = 2(3n - 1)
Sloan [31]
2(n - 1) + 4n = 2(3n - 1)
2n - 2 + 4n = 6n -2
6n - 2 = 6n - 2 (equal)
6n - 6n = 2 - 2
0 = 0

whatever value you put into the place of n, you'll get 0 = 0
3 0
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
Use the zero product property to find the solutions to the eqation 6x^2 -5x=56​
nignag [31]

Answer:

B. 7/2, -8/3

Step-by-step explanation:

Move all the terms to the left side and set them equal to zero.  Then set each factor equal to zero.

8 0
3 years ago
HELP ASAP!!!!
Alecsey [184]

Answer:

A, C, E, G

Step-by-step explanation:

8 0
3 years ago
The sum of X and fifteen
zhuklara [117]

x+15

depends on the value of x

7 0
3 years ago
Read 2 more answers
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