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Marianna [84]
3 years ago
7

PLEASE HELP!

Mathematics
1 answer:
Kitty [74]3 years ago
8 0
You can reword the two equations as:

-5x-y=15 (Divide original value by 3)
-2x+6y=6

Then use elimination to find x:

-30x-6y=90 (Multiply by 6 to get y values to be same to cancel out)
-2x+6y=6

You're left with:

-32x=96. Which can then be solved to find x which is -3.

Then plug back in 

-2x+6y=6

Now to: -2(-3)+6y=6.

Which reduces to 6+6y=6. So y=0.



To graph them, just reword the equations (yes once again) so that y is in front.

y=-5x-15 and y=(1/3)x+1






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Write each number in word form and short form<br>9) 543,698<br>10) 45,987,159.9​
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8 0
3 years ago
Suppose you can factor x^2 +bx +c as (x+p)(x+q) . If c&lt;0, what could be the possible values of p and q?
Anna11 [10]
Ans: Option A

Explanation:
Let's solve it smartly!
Given expression: x^{2} + bx +c --- (A)
Factors: (x+p)(x+q)
Condition: c<0
Now let us expand (x+p)(x+q):
=> x^{2} + (p+q)x + pq  --- (B)

By comparing (B) with (A), we can say that:
pq = c --- (C)

Now, as the condition says, c<0, it means either p or q is negative. Both cannot be positive or both cannot be negative.

1) If p>0, q>0, it means c>0 since (+p)(+q) = (+c)(according to equation (C)). Condition is not met.
Hence, option B and D are wrong.

2) If p<0, q<0 it means c>=0 since (-p)(-q) = (+c)(according to equation (C)). Condition is not met.
Hence option C is out as well.

We are left with Option A:p<0, q>0 it means c<0 since (-p)(+q) = (-c)(according to equation (C)).
Condition is MET!
Hence,
Ans: Option A: p= -3, q= 7
4 0
3 years ago
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