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3 years ago

Solve 11kx + 13kx = 6 for x. A. X= 1/4k B. X= 4k C. X= 4/k D. X= k/4

Mathematics

2 answers:

mixer [17]3 years ago

8 0

Answer:

Option (a) is correct.

$x=\frac{1}{4k}$

Step-by-step explanation:

Given: Equation 11kx + 13kx = 6

We have to solve for x and choose the correct option from the given options.

Consider the given equation 11kx + 13kx = 6

Taking kx common, we have,

kx (11+13) = 6

Simplify, we have,

24kx = 6

Divide both side by 24, we have,

$kx=\frac{6}{24}$

Simplify, we have,

$kx=\frac{1}{4}$

Divide both side by k, we have,

$x=\frac{1}{4k}$

Thus, Option (a) is correct.

rjkz [21]3 years ago

3 0

24kx = 6
kx = 6/24
kx = 1/4
x = 1/4/k
x = 1/4k

Answer: A) X = 1/4k or the first option.

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2 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

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x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

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Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

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2 years ago

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Ray Of Light [21]

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Step-by-step explanation:

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