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Afina-wow [57]
3 years ago
11

Find x- and y-intercepts. Write ordered pairs representing the points where the line crosses the axes. 2x+3y=6

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
8 0

Answer:

x-intercept: (3, 0)

y-intercept: (0, 2)

Step-by-step explanation:

For a function like:

y = f(x)

The x-intercept is the value of x when y = 0

the y-intercept is the value of y when x = 0

Also remember that an ordered pair is written as (x, y).

In this case we have the equation:

2*x + 3*y = 6

For the x-intercept, we just replace y by zero in the equation, then we get:

2*x + 3*0 = 6

Solving this for x, we get:

2*x = 6

x = 6/2 = 3

Then in this case, the ordered pair for the x-intercept is (3, 0)

For the y-intercept, we just need to replace x by zero in the equation:

2*0 + 3*y = 6

Solving this for y, we get:

3*y = 6

y = 6/3 = 2

Then the ordered pair for the y-intercept is (0, 2)

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First you want to distribute the equation.

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Now we add like terms so it comes out to

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Two expressions that can come out to equal 11x + 25 is. 9(x+3) + 2(x+3) or 10(x+2) +(x+5)

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use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 )​, (15 comma
lyudmila [28]

Answer:

volume V of the solid

\boxed{V=\displaystyle\frac{125\pi}{12}}

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.  

Each slice is a quarter of cylinder 5/n thick and has a radius of  

-k(5/n) + 5  for each k = 1,2,..., n (see picture)

So the volume of each slice is  

\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}

for k=1,2,..., n

We then add up the volumes of all these slices

\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}

Notice that the last term of the sum vanishes. After making up the expression a little, we get

\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2

But

\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)

we also know that

\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}

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\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}

so we have, after replacing and simplifying, the sum of the slices equals

\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}

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