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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 figure in the quadrant is rotated 180 counterclockwise about the origin in which quadrant will the rotated figure appear?

abruzzese [7]

Answer: C

C is the correct answer

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3 years ago

Find lim h->0 f(9+h)-f(9)/h if f(x)=x^4 a. 23 b. -2916 c. 2916 d. 2925

Svetach [21]

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h$

Carry out the binomial expansion in the numerator:

$(9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4$

Then the 9⁴ terms cancel each other, so in the limit we have

$\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h$

Since h is approaching 0, that means h ≠ 0, so we can cancel the common factor of h in both numerator and denominator:

$\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)$

Then when h converges to 0, each remaining term containing h goes to 0, leaving you with

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}$

or choice C.

Alternatively, you can recognize the given limit as the derivative of f(x) at x = 9:

$f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h$

We have f(x) = x ⁴, so f '(x) = 4x ³, and evaluating this at x = 9 gives the same result, 2916.

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3 years ago

How would I solve 6^10/6^4 ?

erma4kov [3.2K]

Multiply 6 ten time by itself
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3 years ago

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sineoko [7]

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2 years ago

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3 years ago