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Rama09 [41]
3 years ago
7

What is (5x-2)(4x^2 -3x-2)

Mathematics
2 answers:
Ronch [10]3 years ago
8 0
Answer: 20x^3 - 23x^2 - 4x + 4

Explanation:

Use distributive property:

(5x-2)(4x^2 - 3x-2)
= 20x^3 - 15x^2 - 10x - 8x^2 + 6x + 4
= 20x^3 - 23x^2 - 4x + 4
Novay_Z [31]3 years ago
8 0

Answer:

20x^3-23X^2-4X+4

Step-by-step explanation:

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A) Compute the sum
avanturin [10]
A)

To calculate this sum, we could use trigonometric identity:

\arcsin(x)-\arcsin(y)=\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)

We have:

\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k+1-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{(k+1)^2-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\


=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\dfrac{\sqrt{(k+1)^2-1}}{\sqrt{(k+1)^2}}-\dfrac{1}{k+1}\cdot\dfrac{\sqrt{k^2-1}}{\sqrt{k^2}}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{\dfrac{(k+1)&^2-1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{\dfrac{k^2-1}{k^2}}\right]=\\\\\\=
\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\dfrac{1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{1-\dfrac{1}{k^2}}\right]=\\\\\\


=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\left(\dfrac{1}{k+1}\right)^2}-\dfrac{1}{k+1}\cdot\sqrt{1-\left(\dfrac{1}{k}\right)^2}\right]=\\\\\\=
\sum\limits_{k=1}^n\left[\arcsin\left(\dfrac{1}{k}\right)-\arcsin\left(\dfrac{1}{k+1}\right)\right]=\\\\\\

=\bigg[\arcsin(1)-\arcsin\left(\frac{1}{2}\right)\bigg]+\bigg[\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)\bigg]+\\\\\\+
\bigg[\arcsin\left(\frac{1}{3}\right)-\arcsin\left(\frac{1}{4}\right)\bigg]+\dots+
\bigg[\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)\bigg]=\\\\\\

=\arcsin(1)-\arcsin\left(\frac{1}{2}\right)+\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{1}{3}\right)-\\\\\\-\arcsin\left(\frac{1}{4}\right)+\dots+\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)=\\\\\\=
\arcsin(1)-\arcsin\left(\frac{1}{n+1}\right)=\dfrac{\pi}{2}-\arcsin\left(\frac{1}{n+1}\right)

So the answer is:

\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)}

B)

\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=
\lim\limits_{n\to\infty}\Bigg(\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)\Bigg)=\dfrac{\pi}{2}-\lim\limits_{n\to\infty}\arcsin\left(\dfrac{1}{n+1}\right)=\\\\\\=
\Bigg\{\dfrac{1}{n+1}\xrightarrow{n\to\infty}0\Bigg\}=\dfrac{\pi}{2}-\arcsin(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}

So we prove that:

\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}
7 0
3 years ago
valerie has 10 Video games. 7 of her video games are racing games. what is valerie's racing games written as a decimal?​
svet-max [94.6K]

Answer:

7/10.

Step-by-step explanation:

If theres a total of 10 games and 7 out of the 10 are racing games, it means that 7/10 of the games are racing games.

6 0
3 years ago
Read 2 more answers
E value of y varies directly with x, and y=8 when x=-32. find y when x=64.
ikadub [295]
Y = kx
8 = -32k
k = -8/32 = -1/4
y = -1/4 (64)
y = -16
6 0
3 years ago
Alexa has $60 already saved and her sister Susan has $120 already saved. Alexa is saving $7 per week and her sister is saving $5
masha68 [24]

Step 1:

The variable is the number of weeks needed to equalize Alexa and her sisters savings.

Step 2:

7x + 60 = 5x + 120

Step 3:

7x+60=5x+120

Subtract 60 from both sides.

7x+(60-60)=5x+(120-60)

7x=5x+60

Subtract 5x from both sides

7x-5x=5x+60-5x

2x=60

Divide both sides by 2.

2x/2 = 60/2

x=30

Step 4:

7(30) + 60 = 5(30)+ 120

210+60=150+120

270 = 270

Step 5:

To find the unknown amount of weeks, you need to rewrite the equation isolate x on the left-hand side of the equation. I just subtracting/simplified the equation until I could solve it.

(tbh got stuck on this one)

6 0
3 years ago
12. Make a box-and-whisker plot of the data. (1 point)
Sophie [7]

The box-and-whisker plot of the data, showing the five-number summary, is shown in the image atatched below.

<h3>What is the Box-and-whisker Plot?</h3>

The box-and-whisker plot is a plot that displays the minimum, maximum, median, lower and upper quartile of a data, which is termed as the five-number summary of a data.

Given the data, 20, 23, 28, 14, 13, 24, 18, 11, find the five-number summary:

  • Minimum - 11
  • Lower Quartile - 13.5
  • Median - 19
  • Upper Qaurtile - 23.5
  • Maximum - 28

Therefore, the box-and-whisker plot of the data, showing the five-number summary, is shown in the image atatched below.

Learn more about box-and-whisker ploton:

brainly.com/question/12343132

4 0
2 years ago
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