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2 years ago

Does anyone know the answer?

Mathematics

1 answer:

kolezko [41]2 years ago

8 0

Answer:

Part A: 9.75x > 195

Part B: x > 20

Step-by-step explanation:

Let's say Mark works for x hours. For each hour he works, 9.75 dollars are added to Mark's salary. Therefore, for 1 hour, he gets $9.75, for 2 hours, he gets $9.75 + $9.75 = $9.75 * 2 = $19.50, and for x hours, he gets $9.75 * x as his salary.

We want his salary to be more than $195, so we have $9.75 * x (his salary) > $195, which can be written as 9.75x > 195

divide both sides by 9.75 to solve for x

x > 20

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3 years ago

Read 2 more answers

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PSYCHO15rus [73]

1. 1/9 since you take all the "a"s in the words which is 2/18
2.2/9 same thing as above
3.18/18 since there is no "O" it's a trick question
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6.1/2 and 1/2 coins are always 1/2 unless you flip twice
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3 years ago

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Rina8888 [55]

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3 years ago

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How do you do this question?

Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

$\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}$

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

$V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\$

$\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)$

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

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3 years ago