What are the x-intercept and y-intercept of the graph of y=1/3 x−6 x: Y:

(02.01 LC)

ExtremeBDS [4]

Answer:

Yes it is a function

Step-by-step explanation:

We have to check the ordered pairs to find out if given relation is a function or not.

In an ordered pair, the first element represents the input and the second element represents the output.

The set of inputs is domain and output is range.

For a relation to be function, there should be no repetition in domain i.e there should be unique pairs of input and output.

In the given relation, the domain is {3,5,-1,-2}.

No element is repeated hence it is a function ..

7 0

2 years ago

Please help, very hard question (for me), Will give Brainliest!!!

AleksAgata [21]

The answer is D. This is because when you take 10.65 and subtract 9.95 from it you get 0.07. If you take this number and continuously add to 9.95 you get the numbers provided in the chart. We'll also notice that the 0.07 is only added after 10 minutes of talking (10,20,30,etc.).

8 0

3 years ago

Help ASAP please!!!

masya89 [10]

Total revenue the previous year would be $46,355

7 0

2 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

Simplify the following expression: 4(x+10)-3x-2-x

Ilia_Sergeevich [38]

4x+40-3x-2-x
(4x-3x-x)=4x-4x=0
(40-2)=38
The answer will be 38

6 0

3 years ago