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2 years ago

Please help. This is what I need help with.

Mathematics

1 answer:

Ksivusya [100]2 years ago

6 0

Answer:

I doing it for points 20

Step-by-step explanation:

Im doing it for points

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Find the gcf of 200 and 72

Aleks04 [339]

To get the Greates Common Factor (GCF) of 72 and 200 we need to factor each value first and then we choose all the copies of factors and multiply them:

72: 22233 200: 222 55GCF: 222

The Greatest Common Factor (GCF) is: 2 x 2 x 2 = 8

Answer: 8

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3 years ago

What is the equation of the line that has a slope of 1/5 and a y-intercept of -1

DiKsa [7]

Answer:

C y = 1/5x - 1

Step-by-step explanation:

Slope intercept form: y = mx + b where m = slope and b = y-intercept.

Substitute into the slope intercept form m= 1/5 and b = -1

y = 1/5x - 1

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2 years ago

3(x+y)=y if (x y) is a solution to the equation above and y=0 what is the ratio x/y

Alona [7]

The equation can be rewritten as x = -2/3 y, so x/y = -2/3
However, for y=0, the ratio is undefined as you would be dividing by 0??

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3 years ago

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nordsb [41]

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139

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6 0

2 years ago

Read 2 more answers

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

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

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

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

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago