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Jobisdone [24]
3 years ago
9

Plz Help Plz...this is hard cause im a 7th grader doing 8th grade math plz Help​

Mathematics
1 answer:
Gnesinka [82]3 years ago
5 0
Your answer to this question will be D.
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When you find yourself in an accident your MEDICAL expenses are paid by the car insurance.
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Please help me work this problem
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Answer:

13a

Step-by-step explanation:

9a + 3a + a = 13a

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Read 2 more answers
Sole the system by elimination -4x+9y=9 x-3y=-6
Natalka [10]

Answer:

A) -4x + 9y = 9

B) x -3y = -6

We multiply B) by 3

B) 3x -9y = -18 Then add it to A)

A) -4x + 9y = 9

-x = -9

x = 9

9 -3y = -6

3y = 15

y = 5


Step-by-step explanation:


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3 years ago
Which statement below is correct​
Lynna [10]
The correct answer is A
4 0
3 years ago
The half life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
Veronika [31]

Using an exponential function, it is found that:

a) N(t) = 75(0.5)^{\frac{t}{3.8}}

b) 37.5 grams of the gas remains after 3.8 days.

c) The amount remaining will be of 10 grams after approximately 11 days.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

A(t) = A(0)(1 - r)^t

In which:

  • A(0) is the initial value.
  • r is the decay rate, as a decimal.

Item a:

We start with 75 grams, and then work with a half-life of 3.8 days, hence the amount after t daus is given by:

N(t) = 75(0.5)^{\frac{t}{3.8}}

Item b:

This is N when t = 3.8, hence:

N(t) = 75(0.5)^{\frac{3.8}{3.8}} = 37.5

37.5 grams of the gas remains after 3.8 days.

Item c:

This is t for which N(t) = 10, hence:

N(t) = 75(0.5)^{\frac{t}{3.8}}

10 = 75(0.5)^{\frac{t}{3.8}}

(0.5)^{\frac{t}{3.8}} = \frac{10}{75}

\log{(0.5)^{\frac{t}{3.8}}} = \log{\frac{10}{75}}

\frac{t}{3.8}\log{0.5} = \log{\frac{10}{75}}

t = 3.8\frac{\log{\frac{10}{75}}}{\log{0.5}}

t \approx 11

The amount remaining will be of 10 grams after approximately 11 days.

More can be learned about exponential functions at brainly.com/question/25537936

4 0
2 years ago
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