H (x) = -x+4 when h(x)=-3

Anastaziya [24]

its bella saves more per chore than sweet tea, because 3.5 x c = the amount of chores bella does. Add 11 to that, and she will get more than Sweet Tea. for Sweet tea, one chore saves her twelve dollars. For Bella, one chore saves her more.

8 0

3 years ago

what are some good Netflix show's to watch and don't say all American because i finished both seasons in a week :(..lol

Anit [1.1K]

I love Money heist. I recommend it :)

3 0

3 years ago

Read 2 more answers

Ravi needs to buy some pencils. Brand A has 36 pencils for $8.52. Brand B has a pack of 48 pencils for $9.98. Find the unit pric

Aliun [14]

For this question you would divide $8.52 by 36 pencils to get about $0.24 per pencil for brand A rounded to the nearest cent. Then divide $9.98 by 48 pencils to get about $0.21 per pencil for brand B rounded to the nearest cent.

6 0

3 years ago

Round 392 to the nearest ten

Bezzdna [24]

Nearest 10 is according to the last digit.
if the digit is >5, it +1 to the tenth digit. If it's <5, it will be reduced to 0.

for your case, 392, 2 < 5, therefore rounded up to nearest tenth is 390.

5 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.

5 0

3 years ago