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Harman [31]
3 years ago
5

Solve the simultaneous equations8x – 3y = 303x + y = 7​

Mathematics
1 answer:
hodyreva [135]3 years ago
6 0
Solve for y in 3x+7=y

y=7−3x

Substitute y=7−3x

y=7−3x into 8x−3y=30

8x−3y=30.

17x−21=30

Solve for x

x in 17x−21=30

17x−21=30.

X=3

Substitute x=3 into y=7−3x

y=-2

Therefore,

x=3

y=−2

​

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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
Max and Sasha exercise a total of 20 hours each week. Max exercises 15 hours less than 4 times the
navik [9.2K]

Answer:

Max exercises 13 hours a week, and Sasha 7.

Step-by-step explanation:

To find the number of hours each of them exercises during the week, we solve the system of equations.

In the second equation:

x = 4y - 15

Replacing in the first equation:

x + y = 20

4y - 15 + y = 20

5y = 35

y = \frac{35}{5}

y = 7

So Sasha exercises 7 hours per week.

Max:

x + y = 20

x + 7 = 20

x = 13

Max exercises 13 hours a week.

6 0
3 years ago
Find the length of the third side. If necessary, round to the nearest tenth.This is on Pythagorean theory​
kenny6666 [7]

Answer:

23 because the 24 side is a little bigger than the other side

Step-by-step explanation:

5 0
3 years ago
Lea,Paul and Tynos invest in a business in the ratio 2:3:5.if the profit from the business is $450 000.how much does each person
Digiron [165]

Answer:

see explanation

Step-by-step explanation:

Sum the parts of the ratio, 2 + 3 + 5 = 10 parts

Divide the profit by 10 to find the value of one part of the ratio.

$450 000 ÷ 10 = $45 000 ← value of 1 part of the ratio, then

2 parts = 2 × $45 000 = $90 000 ← Lea's share

3 parts = 3 × $45 000 = $135 000 ← Paul's share

5 parts = 5 × $45 000 = $225 000 ← Tynos' share

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