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stira [4]
3 years ago
5

Y=2x+10 find y if x = -2

Mathematics
2 answers:
Yuki888 [10]3 years ago
8 0
Y=2(-2)+10
Y=-4+10
Y=6
Alexxandr [17]3 years ago
4 0

Answer:

the answer will be y = 6

hope this helps :)

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asambeis [7]

Answer:

4x + 5 = 2x + 7

4x + 5 - 2x = 2x - 2x +7

2x + 5 = 7

2x + 5 - 5 = 7 - 5

2x = 2

2x÷2 = 2÷2

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2 years ago
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12 2/5 x 3 1/6 pls help me
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1. Turn improper
12 \frac{2}{5}  =  \frac{62}{5}
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3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
DedPeter [7]

Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

-30r = \ln{0.5}

r = \frac{\ln{0.5}}{-30}

r = 0.023

So

Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Q(t) = 180e^{-0.023t}

1 = 180e^{-0.023t}

e^{-0.023t} = \frac{1}{180}

e^{-0.023t} = 0.00556

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\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

t = 225.8

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8 0
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X = 4*8 = 32

Then the diagram represents the division:

32/4 = 8 or " 32 divided by 4 = 8"

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