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3 years ago

What is a correct expansion of (3x + 2)(3x^2 + 4)

Mathematics

2 answers:

Anuta_ua [19.1K]3 years ago

5 0

All we need to do is use FOIL here to multiply these.

(3x + 2)(3x^2 + 4)

9x^3 + 12x + 6x^2 + 8

Simplify.

Rearrange terms.

9x^3 + 6x^2 + 12x + 8 is the expanded form of the original binomials.

4vir4ik [10]3 years ago

5 0

The answer is: 3x·3x^2+3x·4+2·3x^2+2·4 APEX

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What are the coordinates of the endpoints of the midsegment for △XYZ that is parallel to XZ⎯⎯⎯⎯⎯ ?

Ksenya-84 [330]

The coordinates of the endpoints of the mid-segment of Δ XYZ that parallel to XZ are (1 , 6) and (1 , 3)

Step-by-step explanation:

In a triangle the segment which joining the mid points of two sides:

Parallel to the 3rd side
Its length is equal half the length of the 3rd side
It's called the mid-segment of the triangle

In Δ XYZ

∵ Vertex X is (0 , 7)

∵ Vertex Y is (2 , 5)

∵ Vertex Z is (0 , 1)

∵ The mid-segment of Δ XYZ is parallel to XZ

- The mid-segment intersects the other two sides of the Δ

at their mid-points

∴ The mid-segment intersects XY and YZ at their midpoints

∴ The endpoints of the mid-segment are the midpoints of

XY and YZ

The mid point of a segments whose endpoints are $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

∵ X = (0 , 7) and Y = (2 , 5)

∴ $x_{1}$ = 0 and $x_{2}$ = 2

∴ $y_{1}$ = 7 and $y_{2}$ = 5

∵ The mid point of XY = $(\frac{0+2}{2},\frac{7+5}{2})$

∴ The mid point of XY = $(\frac{2}{2},\frac{12}{2})$

∴ The mid point of XY = (1 , 6)

∵ Z = (0 , 1) and Y = (2 , 5)

∴ $x_{1}$ = 0 and $x_{2}$ = 2

∴ $y_{1}$ = 1 and $y_{2}$ = 5

∵ The mid point of ZY = $(\frac{0+2}{2},\frac{1+5}{2})$

∴ The mid point of ZY = $(\frac{2}{2},\frac{6}{2})$

∴ The mid point of ZY = (1 , 3)

∵ The endpoints of the mid-segment are the midpoints of

XY and YZ

∴ The end points of the mid segments are (1 , 6) and (1 , 3)

The coordinates of the endpoints of the mid-segment of Δ XYZ that parallel to XZ are (1 , 6) and (1 , 3)

Learn more:

You can learn more about the mid-point in brainly.com/question/5223123

#LearnwithBrainly

7 0

2 years ago

....................... Plz

vesna_86 [32]

Answer:

2/5 (x-1)=15 (x=7)

7 (1+3m)=49 (m=2)

6 0

2 years ago

Read 2 more answers

This is a continuous set of points in the coordinate plane. It can be, but is not always, straight.

castortr0y [4]

Answer:

Linear. While all linear equations produce straight lines when graphed, not all linear equations produce linear functions. In order to be a linear function, a graph must be both linear (a straight line) and a function (matching each x-value to only one y-value).

5 0

2 years ago

[3x^(2/3)] • [7x^(1/4)]<br><br> photo of equation above

aalyn [17]

$\bf a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n}
\qquad \qquad
\sqrt[ m]{a^ n}\implies a^{\frac{ n}{ m}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( 3x^{\frac{2}{3}} \right)\left( 7x^{\frac{1}{4}} \right)\implies 3\cdot 7x^{\frac{2}{3}}\cdot x^{\frac{1}{4}}\implies 21x^{\frac{2}{3}+\frac{1}{4}}\implies 21x^{\frac{(4)2+(3)1}{12}}
\\\\\\
21x^{\frac{11}{12}}\implies 21\sqrt[12]{x^{11}}$

6 0

3 years ago

Read 2 more answers

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

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

and

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

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

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

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

$\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}$

and the volume V of our solid is

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

3 0

2 years ago