Hich values are solutions to the inequality?

Mathematics

2 answers:

Troyanec [42]3 years ago

7 0

$2r\leq3r-8\\0\leq r-8\\8\leq r\\r\geq8$

So, in the list of numbers, 8 and 9 are solutions

Ede4ka [16]3 years ago

7 0

Answer:

8 and 9

Step-by-step explanation:

1. subtract 3r from both sides to get -1r≤-8

2. divide both sides by -1 to get r by itself r=8

3. any number greater than 8 will be a true statement

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Pls help me!! I just need help on both of the following questions, a-d on each

Ymorist [56]

I'll do a and e on the first one to show you how to do it

Then See if you can do the rest

a. 6(-3)

When your multiplying or dividing with negative numbers; If one of your numbers is negative like the one above, your answer will be negative

So 6 × 3 = 18 ; but since there is a negative on the 3 your answer will be -18

e. -20 ÷ 5

same for this; 20 ÷ 5 = 4, but since the 20 is negative the answer will be -4

You do b, c, d, and f on the first question. Tell me what you get and we'll go from there

Now for the second question 

I will do a and d to show you how to do this then you do the rest (I'm inserting a picture with these)

a. $4(-2 \frac{1}{3}) + 7 \frac{1}{6}$
First go ahead and turn the fractions into improper fractions, this makes them a bit easier to work with. When multiplying fractions; they do not need a common denominator, all you do is multiply top by top and bottom by bottom.
However, when you add and subtract fractions you MUST have a common denominator

d. (-4.25)(2) + (-4.25)(-2)
remember from the first problem a negative times a positive gives you a negative

but when you times a negative by a negative you get a positive

Now you try b and c; let me know what you get, and we'll go from there!