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2 years ago

Can i get help pldssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss

Mathematics

1 answer:

Ghella [55]2 years ago

8 0

A: no
X = 3
2x + 3
2(3) + 3
6 + 3
9
For the other one:
2(x + 3)
2(3+3)
2(6)
12

2x + 3 and 2(x + 3) are not equivalent because one has parentheses, which means you will have to do parentheses first instead of multiplication (PEMDAS rule)
Hope this helps!

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one vertex of a triangle is located at (0,5) on a coordinate grid after a transformation the vertex is located at(5,0) which tra

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Answer:

It's a rotation clockwise of 90 degrees about the origin.

Step-by-step explanation:

The given point moves from y = 5 to x = 5 so it has passed through 90 degrees about the origin.

It's a rotation clockwise of 90 degrees about the origin.

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$-\frac{6}{7}$

Step-by-step explanation:

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1 year ago

Evan walks his dog 4 times around the perimeter of a

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2 years ago

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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2 years ago