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3 years ago

8. Find the value of k, if (1-1) is a solution of the equation 3x-ky =8. Also, find the coordinates of

Mathematics

1 answer:

Yakvenalex [24]3 years ago

8 0

Answer:

k = 5 and (6,2).

Step-by-step explanation:

Since (1,-1) is a solution of the equation 3x - ky = 8, so the point (1,-1) will satisfy the equation above.

Hence, putting x = 1 and y = -1 in the equation will give left-hand side = right-hand side.

So, 3(1) - k(-1) = 8

⇒ 3 + k = 8

⇒ k = 5 (Answer)

Therefore, the equation of the straight line is 3x - 5y = 8 ....... (1)

Now, putting x = 6 , then from equation (1) we get y = 2

Therefore, (6,2) is also a point on the graph of equation (1). (Answer)

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A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$

Which gives;

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2022} \right)$

$\displaystyle A = 2 \times 1011 \cdot \left(1 - \frac{1}{2022} \right) = \frac{1032231}{511} \approx \mathbf{2020.022}$

A ≈ <u>2020.022</u>

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