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shtirl [24]
3 years ago
11

Can you show work for all of these and match them

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
4 0

I'll show you how to do one of the equations and one of the inequalities. All the others are done exactly in the same way: you'll only have to change the numbers, and it will be a good exercise.

Equations

Let's take the first equality as an example: we have

|x-2|=5

By definition, the absolute value of a number is the positive version of that number: if the number is already positive the absolute value doesn't change it; if a number is negative the absolute value changes its sign.

So, if the absolute value of a number is 5, than that number was already 5, or it was -5, and the absolute value changed it to positive 5.

So, the solutions are given by

|x-2|=5 \iff x-2=5 \lor x-2 = -5 \iff x=7 \lor x=-3

Inequalities

Again, we'll use the first one as example. We have

|x+4|\geq 7

By the same logic as before, the absolute value of a number is greater than 7 if the number is already greater than 7, or if it is smaller than -7. For example, we have |-10|=10>7.

So, we have

|x+4|\geq 7 \iff x+4 \geq 7 \lor x+4 \leq -7 \iff x \geq 3 \lor x \leq -11

Instead, if we have an inequality with the "less than" sign, we have for example

|x-5|\leq 8 \iff -8 \leq x-5 \leq 8 \iff -3 \leq x \leq 13

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\frac{3}{7}*r + \frac{5}{8} *s

let us first plug the values of r and s here

r=14 and s=8

\frac{3}{7}*14 + \frac{5}{8} *8

Now here we can cancel 14 and 7 as 7*2 =14 , also we can cancel 8 and 8 as 8*1 is 8, so our simplified new form is:

\frac{3}{1}*2 + \frac{5}{1} *1

Now we can multiply 3*2 and 5*1

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Next we add the two numbers

6+5 =11

So this is how we get the answer 11

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