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svetlana [45]
3 years ago
14

Solve and graph y>3x y>-x-8

Mathematics
1 answer:
OleMash [197]3 years ago
4 0

Answer: it’s says the equation is solved here’s the graph

Step-by-step explanation:

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Please help!!!!!!!!!!!!!!!!
riadik2000 [5.3K]
Write f(x) / g(x) = 1 / ((x+1)(x-2)).

It's easy to see that x cannot be -1 and 2;
So all real numbers exclude -1 and 2, which matches the second choice.
6 0
3 years ago
PLEEZZ
jonny [76]

4 < 4(6y - 12) - 2y       |use distributive property

4 < (4)(6y) + (4)(-12) - 2y

4 < 24y - 48 - 2y       |add 48 to both sides

52 < 26y      |divide both sides by 26

2 < y → y > 2

---------------------------------------------------------

4x + 3 < 3x + 6      |subtract 3 from both sides

4x < 3x + 3       |subtract 3x from both sides

x < 3

------------------------------------------------------------

-5r + 6 ≤ -5(r + 2)        |use distributive property

-5r + 6 ≤ (-5)(r) + (-5)(2)

-5r + 6 ≤ -5r - 10         |add 5r to both sides

6 ≤ -10     FALSE

No solution

-------------------------------------------------------------

-2(6 + s) ≥ -15 - 2s        |use distributive property

(-2)(6) + (-2)(s) ≥ -15 - 2s

-12 - 2s ≥ -15 - 2s        |add 2s to both sides

-12 ≥ -15     TRUE

All real numbers

-------------------------------------------------------------

3s + 6 ≤ -5(s + 2)        |use distributive property

3s + 6 ≤ (-5)(s) + (-5)(2)

3s + 6 ≤ -5s - 10         |subtract 6 from both sides

3s ≤ -5s - 16        |add 5s to both sides

8s ≤ -16         |divide both sides by 8

s ≤ -2

--------------------------------------------------------------

4/3 s - 3 ?

4 0
3 years ago
How many possible solutions could f(x) = 6x^3 + 8x^2 - 7x -3 have?
Alex17521 [72]
3 possible solutions.


The reason is because of the degree, which is 3
3 0
3 years ago
Read 2 more answers
Critique Reasoning James says is greater than 10. Is James correct? Explain.​
Dafna1 [17]
The is answe is less than because I have to get them up
3 0
2 years ago
Verify a(b-c)=ab-ac for a=1.6;b=1/-2;&amp; c=-5/-7​
harina [27]

Given:

a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}

To verify:

a(b-c)=ab-ac for the given values.

Solution:

We have,

a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}

We need to verify a(b-c)=ab-ac.

Taking left hand side, we get

a(b-c)=1.6\left(\dfrac{1}{-2}-\dfrac{-5}{-7}\right)

a(b-c)=1.6\left(-\dfrac{1}{2}-\dfrac{5}{7}\right)

Taking LCM, we get

a(b-c)=1.6\left(\dfrac{-7-10}{14}\right)

a(b-c)=\dfrac{16}{10}\left(\dfrac{-17}{14}\right)

a(b-c)=\dfrac{8}{5}\left(\dfrac{-17}{14}\right)

a(b-c)=-\dfrac{68}{35}\right)

Taking right hand side, we get

ab-ac=1.6\times \dfrac{1}{-2}-1.6\times \dfrac{-5}{-7}

ab-ac=-\dfrac{16}{20}-\dfrac{8}{7}

ab-ac=-\dfrac{4}{5}-\dfrac{8}{7}

Taking LCM, we get

ab-ac=\dfrac{-28-40}{35}

ab-ac=\dfrac{-68}{35}

Now,

LHS=RHS

Hence proved.

7 0
2 years ago
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