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Ivan
4 years ago
12

Find two polynomials that have a sum of 4

Mathematics
1 answer:
Furkat [3]4 years ago
4 0
There are infinitely many ways to do this. One such way is 
(x^2 + 10x + 4) plus (-x^2 - 10x)
note how the x^2 terms combine to 0x^2 or just 0. The same applies to the x terms as well. The only thing left is 4

So as you can see, the goal is to get everything to cancel but the 4. You can also have something like 
(2x^2-7x+2) plus (-2x^2 + 7x + 2)
and we have the same cancellations going on. This time we have 2+2 = 4 left over. 
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Wewaii [24]
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6 0
3 years ago
I WILL MARK YOU AS BRAINLIEST!PLEASE HELP ME WITH 19
Alex787 [66]

Answer:

x is 7.5 so the answer is D

Step-by-step explanation:

Since the triangles are similar, you can flip the 1st one to match the 2nd.

Now you can write

16/12=10/x

(the 9 is irrelevant)

now cross multiply to solve for x

12x10=120

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3 0
3 years ago
HIGH POINTS)Please help! (20 points)
USPshnik [31]

Answer:

What the other person said

Step-by-step explanation:

4 0
3 years ago
Describe how multiplying all the linear dimensions of a cone by one
Anton [14]

Answer:\frac{1}{64}^{th}

Step-by-step explanation:

Given

Suppose V is the original volume of cone having r as radius and h as height

If we multiply all the linear dimensions of ac one by \frac{1}{4}^{th}

the r'=\frac{r}{4}

h'=\frac{h}{4}

Therefore new volume is

V'=\frac{1}{3}\pi r'^2h'

V'=\frac{1}{3}\pi (\frac{r}{4})^2(\frac{h}{4})

V'=\frac{1}{64}\times \frac{1}{3}\pi r^2h

V'=\frac{V}{64}

So new volume becomes \frac{1}{64}^{th} of the original one

6 0
3 years ago
What is the mode: 18,21,22,18,19
Evgesh-ka [11]

Answer:

18

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Mode =  modal class (Recurring)
The number/value that appears the most which in this case is 18 because it can be seen twice.

6 0
2 years ago
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