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Nimfa-mama [501]
2 years ago
14

What is the range of this graph?

Mathematics
2 answers:
torisob [31]2 years ago
4 0
Can’t see the graph, try reposting it
Lelechka [254]2 years ago
3 0
Did you forgot to put your picture in the question
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Write a ratio for the situation in three ways, comparing the first quantity to the second quantity. In Largo City, about one in
saul85 [17]

Answer:


Step-by-step explanation:

1/9

(1*4) / (9*4) = 4/36

(1*4) / (9*4) = 5/45

The first one is the ratio as it is stated in the problem.

The second one multiplies top and bottom by 4

The third one multiplies top and bottom of the top ratio by 5.

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2 years ago
PLEASE HELP BOTS HAVE BEEN COMMENTING ON MY POSTS ALL DAY I NEED AN REAL ANSWER PLEASE IM DEPERATE
allochka39001 [22]
Yep the same thing happens to me
8 0
3 years ago
Which equation has the same solution as x – 14x + 20 = 3?
zavuch27 [327]

Answer:

x=17

Step-by-step explanation:

3 0
2 years ago
find the domain and range of the relation (-3,3),(-3,2),(-3,1)(-3,0).Then determine whether the relation is a function
navik [9.2K]
The domain is the x value: (-3)
The range are the y values: (0, 1, 2, 3) in order from least to greatest
No this is not a function because that value of x is "reused". You can "reuse" a y value with different x values but you can't reuse an x value ever, even if the y's are different. x's cannot be the same
4 0
3 years ago
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
3 years ago
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