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Gennadij [26K]
3 years ago
6

What is the product (-6a^3b+2ab^2)(5a^2-2ab^2-b)

Mathematics
2 answers:
frutty [35]3 years ago
8 0

-30a^5b+12a^4b^3+16a^3b^2-4a^2b^4-2ab^3

sasho [114]3 years ago
5 0

the answer is <u><em>B</em></u> on edg :)

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First do the Base multiplied by the Height.

After divide by 2

Step-by-step explanation:


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Can anyone help me with this please
Gennadij [26K]

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A

Step-by-step explanation:

the other options are constants, option A is the only option that would just be represented by a whole number

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What is the Slope and y intercept
Inessa [10]

Answer:

The slope is .075 and the y intercept is -.125

Step-by-step explanation:

8y = .2(3x -5)

Distribute the .2

8y = .6x - 1

Divide by 8

8y/8 = .6x/8 -1/8

y = .075x -.125

This is in slope intercept form  y= mx+b  where m is the slope and b is the y intercept.

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3 years ago
Find the length of the third side to the nearest tenth.<br> Hurry need help asap!!
gavmur [86]

Answer:

2.8

Step-by-step explanation:

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3 years ago
In a G.P the difference between the 1st and 5th term is 150, and the difference between the
liubo4ka [24]

Answer:

Either \displaystyle \frac{-1522}{\sqrt{41}} (approximately -238) or \displaystyle \frac{1522}{\sqrt{41}} (approximately 238.)

Step-by-step explanation:

Let a denote the first term of this geometric series, and let r denote the common ratio of this geometric series.

The first five terms of this series would be:

  • a,
  • a\cdot r,
  • a \cdot r^2,
  • a \cdot r^3,
  • a \cdot r^4.

First equation:

a\, r^4 - a = 150.

Second equation:

a\, r^3 - a\, r = 48.

Rewrite and simplify the first equation.

\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}.

Therefore, the first equation becomes:

a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150..

Similarly, rewrite and simplify the second equation:

\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}.

Therefore, the second equation becomes:

a\, r\, \left(r^2 - 1\right) = 48.

Take the quotient between these two equations:

\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}.

Simplify and solve for r:

\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}.

8\, r^2 - 25\, r + 8 = 0.

Either \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16} or \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}.

Assume that \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}.

Similarly, assume that \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = \frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}.

4 0
3 years ago
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