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zzz [600]
2 years ago
6

Someone please answer this for me (Algebra 2)

Mathematics
1 answer:
Tema [17]2 years ago
6 0

Using division of fractions, the equivalent expression is:

B. \frac{4}{d - 6}.

<h3>How to divide fractions?</h3>

To divide two fractions, we multiply the first by the inverse of the second.

In this problem, we have that:

  • The first fraction is: \frac{2d - 6}{d^2 + 2d - 48} = \frac{2(d - 3)}{(d + 8)(d - 6)}.
  • The second fraction is: \frac{d - 3}{2d + 16} = \frac{d - 3}{2(d + 8)}.

Applying the multiplication, we have that:

\frac{2(d - 3)}{(d + 8)(d - 6)} \times \frac{2(d + 8)}{d - 3} = \frac{4}{d - 6}.

Hence option B is correct.

More can be learned about division of fractions at brainly.com/question/8433134

#SPJ1

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Jermaine did this work to solve an equation. Did he make an error?
Nata [24]

The solution for the below equation is required:

4x + 6 - x = 2x + 3

The work done by the Jermaine is presented below :

4x + 6 - x = 2x + 3

5x + 6 = 2x + 3

Instead of adding 4x with -1x to make it 3x , he added 4x with 1x to make it 5x.

So, the correct option is

Jermaine did make an error. He should have combined 4x and -1x to make 3x

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6 0
4 years ago
Supose that the half-life of a cesium is 30 years. If there were initially 1000g of the substance,
AveGali [126]

After 200 years, 9.84 grams of Cesium would remain.

Let N represent the amount of substance after t years

The half-life of a cesium is 30 years., hence, this can be represented by exponential function:

N(t)=ab^t

where a is the initial value and b is the multiplier.

There were initially 1000g of the substance

a) The exponential model for this situation is:

N(t) =1000(\frac{1}{2} )^{\frac{1}{30} t}\\\\N(t) =1000(\frac{1}{2} )^{\frac{t}{30}

b) After 200 years (t = 200), the amount of Cesium remaining is:

N(200)=1000(\frac{1}{2} )^\frac{200}{30} \\\\N(200) = 9.84g

Hence after 200 years, 9.84 grams of Cesium would remain

Find out more at: brainly.com/question/24710827

8 0
3 years ago
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