11kx+13kx= 6 for x a: x=4/k b:x=1/4k c:x=4k d:x=k/4 i have no clue smh

Mathematics

2 answers:

lbvjy [14]2 years ago

8 0

24kx = 6

kx = 6/24

kx = 1/4

x = 1/4/k

x = 1/4k

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

lukranit [14]2 years ago

3 0

Ion have a clue either just guess C

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Lisa [10]

<h3>Answer: 10</h3>

=======================================================

Explanation:

We'll start off converting each mixed number into an improper fraction.

The formula to use is $A \frac{b}{c} = \frac{A*c+b}{c}$

So,

$A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{1}{8} = \frac{2*8+1}{8}\\\\2 \frac{1}{8} = \frac{16+1}{8}\\\\2 \frac{1}{8} = \frac{17}{8}\\\\$

And,

$A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{2}{3} = \frac{2*3+2}{3}\\\\2 \frac{2}{3} = \frac{6+2}{3}\\\\2 \frac{2}{3} = \frac{8}{3}\\\\$

So the task of computing $2 \frac{1}{8} \times 2 \frac{2}{3}$ is exactly the same as computing $\frac{17}{8} \times \frac{8}{3}$

Notice how we have an 8 up top and an 8 down below. Those 8's cancel out and we're left with $\frac{17}{3}$ . That fraction does not reduce any further.