Pls answer this pls answer this pls answer this pls answer this pls answer this pls answer this

Roman55 [17]

Percent increase = ((new number - original number) / (original number)) x 100.

percent increase = ((3000 - 2500) / 2500)) x 100 =
(500 / 2500)....x 100 =
0.2 x 100 =
20% increase <====

7 0

3 years ago

Please help me with those question please help with out number 1

alina1380 [7]

Answer:

1) For each value of x, a value of y is increased by 5.

x = 0, y = 5

x = 1, y = 10

x = 2, y = 15

x = 3, y = 20

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2) For each two values of x, a value of y is increased by 10.

x = 0, y = -2

x = 2, y = 8

x = 4, y = 18

x = 6, y = 28

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3)

x = 0, y = 1

x = 1, y = $\frac{7}{5}$

x = 5, y = 3

x = 10, y = 5

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4)

x = 0, y = 2

x = 1, y = 17

x = 2, y = 32

x = 5, y = 77

Step-by-step explanation:

This is as easy as replacing x for the actual value show on the table.

1)

$y=5x+5$

When x = 0, y = ?

$y=5(0)+5\\y=0+5\\y=5$

When x = 1, y = ?

$y=5(1)+5\\y=5+5\\y=10$

When x = 2, y = ?

$y=5(2)+5\\y=10+5\\y=15$

When x = 3, y = ?

$y=5(3)+5\\y=15+5\\y=20$

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2)

$y=5x-2$

When x = 0, y = ?

$y=5(0)-2\\y=0-2\\y=-2$

When x = 2, y = ?

$y=5(2)-2\\y=10-2\\y=8$

When x = 4, y = ?

$y=5(4)-2\\y=20-2\\y=18$

When x = 6, y = ?

$y=5(6)-2\\y=30-2\\y=28$

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3)

$y=\frac{2}{5}x +1$

When x = 0, y = ?

$y=\frac{2}{5}(0) +1$

$y=0 +1\\y=1$

When x = 1, y = ?

$y=\frac{2}{5}(1) +1\\y=\frac{2}{5} +1$

$y=\frac{2}{5}+1(\frac{5}{5})\\ y=\frac{2}{5}+\frac{5}{5} \\y=\frac{7}{5}$

When x = 5, y = ?

$y=\frac{2}{5}(5) +1\\y=2+1\\y=3$

When x = 10, y = ?

$y=\frac{2}{5}(10) +1\\y=2(2) +1\\y=4+1\\y=5$

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4)

$y=15x+2$

When x = 0, y = ?

$y=15(0)+2\\y=0+2\\y=2$

When x = 1, y = ?

$y=15(1)+2\\y=15+2\\y=17$

When x = 2, y = ?

$y=15(2)+2\\y=30+2\\y=32$

When x = 5, y = ?

$y=15(5)+2\\y=75+2\\y=77$

4 0

3 years ago

Read 2 more answers

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.