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Paha777 [63]
3 years ago
14

HELP ASAP DOING ALPHA PLUS

Mathematics
1 answer:
ipn [44]3 years ago
3 0

Answer:

D is the answer!!

Step-by-step explanation:

so first you see all the numbers then you see that 0.45 is the least then you look to see and you see 2  problems that have 0,45 then you lokk at the second number there and D has 0.75 and you see that D have least to greatest!!

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Simplify the expression 5(8-m)
labwork [276]

Answer:

40-5m

Step-by-step explanation:

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What is the greatest common factor of 63 45 72 and 36?
trapecia [35]
List factors
63-1,3,7,9,21,63

36-1,2,3,6,12,18,36

45-1,3,5,9,15,45

72-1,2,3,4,6,8,9,12,18,24,36,72

Find biggest number in all of them...
The greatest common factor is 3



8 0
3 years ago
Triangle PQR has been reduced with D as the center of dilation and scale to form triangle P’Q’R’. The figure shows two triangles
julsineya [31]

Answer:

A and B

Step-by-step explanation:

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4 0
3 years ago
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Hey! please help posted picture of question
QveST [7]
C. I am pretty sure this would be correct! I am so sorry if it isn't
5 0
4 years ago
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Let V be the volume of the solid obtained by rotating about the y-axis the region bounded by y=√x and y=x^2. Find V either by sl
Ede4ka [16]

Answer:

The volume of the solid is 3\pi/10

Step-by-step explanation:

In this case, the washer method seems to be easier and thus, it is the one I will use.

Since the rotation is around the y-axis we need to change de dependency of our variables to have f(x)\rightarrow f(y). Thus, our functions with y as independent variable are:

x=\sqrt{y}\\ x=y^2

For the washer method, we need to find the area function, which is given by:

A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]

By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function x=\sqrt{y} and the inner one by x=y^2. Thus, the area function is:

A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)

Now we just need to integrate. The integration limits are easy to find by just solving the equation \sqrt(y)=y^2, which has two solutions y=0 and y=1. These are then, our integration limits.

V=\pi\int_{0}^1 (y-y^4)dy\\ V=\pi (\int_{0}^1 ydy - \int_{0}^1 y^4dy)\\ V=\pi/2-\pi/5\\\boxed{V=3\pi/10}

3 0
4 years ago
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