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2 years ago

Please do a and b and explain

Mathematics

1 answer:

Blizzard [7]2 years ago

4 0

This is a decay problem,

The equation is written as start value x (1 - rate)^tike

Using the information given in the problem:

a) Total = 120(1-0.12)^Hours

b) Replace hours with 19 and solve:

Total = 120(1-0.12)^19

Total = 120(0.88)^19

Total = 10.57674 mg

Round the answer as needed.

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Then find the area S of the region.<br><br> y=2 x^2, y^2=1/2 x

r-ruslan [8.4K]

Recall that the area under a curve and above the x axis can be computed by the definite integral. If we have two curves

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then the area between them bounded by the horizontal lines x = a and x = b is

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2 years ago

If you are selling a item for 38.50 for 30% what would it be

Rzqust [24]

If you were selling an items for $35.50 as the original price, but you give 30% off the item price then the result of it's cost would be:
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3 years ago

Read 2 more answers

PLEASE HELP!

Natasha_Volkova [10]

The graph first represents the provided piecewise function option first is correct.

<h3>What is a piecewise function?</h3>

The graph of a piecewise function contains numerous curve components. It signifies that it has a plethora of definitions based on the value of the input. In other words, a piecewise function behaves differently depending on the input.

We have a piecewise function shown in the picture.

$\rm y=\left\{x\le-2:\sqrt{-x-2}\right\}$

$\rm y=\left\{-2 < x < 1:\dfrac{x}{x^{2}+x-2}\right\}$

$\rm y=\left\{x\ge1:\log_{2}\left(x+1\right)\right\}$

After plotting all the pieces of a function on the coordinate plane we will get a graph the same as shown in the first option.

Thus, the graph first represents the provided piecewise function option first is correct.

Learn more about the piecewise function here:

brainly.com/question/12561612

#SPJ1

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1 year ago

Determine whether the integral is convergent or divergent. [infinity] 7 e−1/x x2 dx convergent divergent Changed: Your submitted

inessss [21]

I suppose the integral could be

$\displaystyle\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx$

In that case, since $-\frac1x\to0$ as $x\to\infty$ , we know $e^{-1/x}\to1$ . We also have $\left(e^{-1/x}\right)'=\frac{e^{-1/x}}{x^2}>0$ , so the integral is approach +1 from below. This tells us that, by comparison,