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anyanavicka [17]
4 years ago
10

Solve the Inequality. -5r + 6 ≤ -5(r+2)

Mathematics
2 answers:
LUCKY_DIMON [66]4 years ago
6 0
There are no real solutions hope this helps :D
Arturiano [62]4 years ago
3 0
There is no solution.
There is no values that make the inequality true, the inequality has no solution.
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How do you solve 30>-(5z+15)+10z
AlekseyPX
Math:

Step #1: Simplify both sides of the inequality.
<span>30><span>5z−15

</span></span>Step #2: Flip the equation.

<span><span>5z−15</span><30

</span>Step #3: Add 15 to both sides.
<span><span><span>5z−15</span>+15</span><<span>30+15

</span></span><span>5z<45

</span>Step #4: Divide both sides by 5.

<span><span>5z/5=45

Your Answer:
</span></span><span>z<9</span>
6 0
3 years ago
Read 2 more answers
Please can’t Gail this
ANEK [815]

Answer:

4

Step-by-step explanation: because you up four and move 1 so the answer is 4/1 but is simplified by 4

3 0
4 years ago
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Can someone sub to my youttube channel pls, its called Yakobu, it would really help out :)
earnstyle [38]

Answer:

yea sure

Step-by-step explanation:

6 0
4 years ago
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Show that (a - b)+(b-c)+(c -a)3 = 3 (a - b) (b -c) (c-a)​
nikdorinn [45]

Answer:

I think that it should be

{(a - b)}^{3}  +  {(b - c)}^{3}  +  {(c - a)}^{3}  = 3(a - b)(b - c)(c - a)

Step-by-step explanation:

Here,

we take , a - b = A,b-c = B , c - a= C

A+B+C = 0

we know that,

{a}^{3}  +  {b}^{3}  +  {c}^{3}   - 3abc = (a + b + c)(  {a}^{2}  +  {b}^{2}  +  {c}^{2}  - ab - bc - ca)

Here , A+B+C = 0

so,

A^3 +B^3 + C^3 = 3 ABC

now we put the values

{(a - b)}^{3}  +  {(b - c)}^{3}  +  {(c - a)}^{3}  = 3(a - b)(b - c)(c - a)

I am done .

4 0
4 years ago
Read 2 more answers
Need Help With Math. One Question
alina1380 [7]

Answer: 2 < x < 16

The triangle inequality theorem says that if we have a triangle with sides a,b,c then

b-a < c < b+a

with b being longer than 'a'

In this case, a = 7 and b = 9 and c = x

so we have

b-a < c < b+a

9-7 < c < 9+7 .... replace a with 7, replace b with 9

2 < c < 16

2 < x < 16 .... replace c with x

So x can be any number between 2 and 16. The value of x cannot be equal to 2. Also, x cannot be equal to 16 either.

8 0
4 years ago
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