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Oksi-84 [34.3K]
3 years ago
5

If a^2+3a+9=0, What is a^3?

Mathematics
1 answer:
kifflom [539]3 years ago
7 0

i dunooo i dont even nooooo

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Alja [10]
The answer should be 7/10!
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Read 2 more answers
Answer? Please ASAP I mark as brainlist please ASAP
Sergio [31]

Answer:

b. (x-3)(x+2)

Step-by-step explanation:

y=x^2-x-6

y=x^2-3x+2x-6

=x(x-3)+2(x-3)

=(x-3)(x+2)

4 0
3 years ago
Plz Help:)
Svetlanka [38]
I think it’s D, hope that helps
8 0
3 years ago
-8(6-r)=24 what is the solution for r
Andreas93 [3]

Answer:hi

Step-by-step explanation:Let me first warn you that I CANNOT get the denominators of this problem to go underneath the numerators. It looks good on my answer page but when I click "preview your answer" the denominators are ALL over the place. Grrrrr.

I think the weirdly placed denominators make this explanation hard to understand but you give this a go and see if you understand things. If it is too hard to see, maybe this website will let you instant message me or email me. If so, then I will send you the solution via email. I'm very sorry the denominators are SO scattered.

See what you think of the explanations below but it might help if you write down my answer on paper, so that you can move the number to rest more closely under their proper numerators. :-)

Ok then......

Let's start with your first problem. A quick way to do this work is to cross multiply. Here's what I mean:

__3___ = __6___

r - 1 r + 2

Ok, so you can cross multiply:

3(r + 2) = 6(r - 1) (cross multiply)

3r + 6 = 6r - 6 (use the distributive property)

Now you will have to put the "r" variables on the left side of the equal sign and numbers without variables on the right side of the equal sign. To get 6r from the right side to the left, you can subtract 6r from the right side. If you subtract it from the right, you must subtract it from the left.....

To get 6 from the left side to the right, you can subtract it from the left. If you subtract it from the left, you must subtract it on the right.....

So now you have:

3r - 6r = -6 - 6 This becomes.........

- 3r = -12

Now to solve for "r" you will have to divide by -3. Why? Well, -3r is really saying: -3 times r. The -3 and the r are actually being multiplied together so to isolate the "r" you have to sort of "undo" the multiplication. What is the opposite of multiplication? It's DIVISION. Soooo, you must divide by -3 on both sides. SO you have:

- 3r = -12

r = 4 (-12 divided by -3 is a positive 4)

Now you know "r" equals 4. If we "plug" 4 back into the original equation, we can see the answer is correct. Check this out:

__3___ = __6___ (original equation)

r - 1 r + 2

__3___ = __6___ (substitute "4" in for "r"

4 - 1 4 + 2

__3___ = __6___ (do the math in the denominator area)

 

3 6

1 = 1 (3/3 equals 1 and 6/6 also equals 1, so our

solution that r = 4 is correct!) :-)

Now to your 2nd problem..........

4/2x-5 = 24/2x

___4___ = __24___

2x � 5 2x

Let's cross multiply again:

4(2x) = 24(2x - 5)

8x = 48x - 120

8x - 48x = - 120

-40x = - 120

(divide both sides by -40)

x = 3

Now let's "plug" 3 in where we had the variable "x" in the original equation to see if we are correct.

___4___ = __24___ (original equation)

2x � 5 2x

___4___ = __24___ (substituted 3 for "x")

2(3) � 5 2(3)

___4___ = __24___

6 - 5 6

___4___ = __24___

1 6

4 = 4

SO once again we have found the correct answer. Yay for us! :-)

I hope this helps. :-)

8 0
3 years ago
Read 2 more answers
Which is bigger: 0.3 or 1/3
valkas [14]
First, turn the fraction into a decimal to get a better look at the problem. To do that, divide the numerator (top half) by the denominator (bottom half). 1 divided by 3 = 0.33
0.33 is larger than 0.3. If it helps, add a 0 to 0.3 to get 0.30 and you'll see that 33 is in fact a larger number than 30. (Adding the zero does NOT change the number, it holds no place value it's just there for looks.)
4 0
3 years ago
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