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OLEGan [10]
3 years ago
7

(2x4+x3−4x2+3)+(4x4−2x3−x2+10x+2)

Mathematics
2 answers:
solniwko [45]3 years ago
8 0
Im assuming there is exponents so if im assuming correctly, then the answer would be 6x^4 - x^3 - 5x^2 + 10x + 5 because you would combine like terms. 
Sidana [21]3 years ago
4 0

Answer:  The required sum is 6x^4-x^3-5x^2+10x+5.

Step-by-step explanation:  We are given to find the following sum:

S=(2x^4+x^3-4x^2+3)+(4x^4-2x^3-x^2+10x+2).

To find the sum, we need to add the coefficients of the same exponents of the unknown variable x.

The summation is as follows:

S\\\\=(2x^4+x^3-4x^2+3)+(4x^4-2x^3-x^2+10x+2)\\\\=(2+4)x^4+(1-2)x^3+(-4-1)x^2+10x+(3+2)\\\\=6x^4+(-1)x^3+(-5)x^2+10x+5\\\\=6x^4-x^3-5x^2+10x+5.

Thus, the required sum is 6x^4-x^3-5x^2+10x+5.

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Slope
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Answer:

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Step-by-step explanation:

This is because during each minute, he does another 2 problems. Or, in other words +2 each minute.

This is the result of the sequence during each minute:

(Problems Left): <em>30, 28, 26, 24, 22... 0</em>

Time (Minutes) --->

If it was instead a geometric equation, it would go by a multiple, such as the number 2. Or, ×2 each minute.

During the first minute, he learns 2 problems. But then as each minutes passes it turns into 4, then 8, and etc..

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Time (Minutes) --->

The formula would be for this equation would follow

an = a1 + (n-1)d

an is your result.

a1 is the first number you plug in

(n-1) is just there, don't mind it (aka: the term position).

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3 years ago
Please help me! I'm having trouble solving this! In the △PQR, PQ = 39 in, PR = 17 in, and the height PN = 15 in. Find QR. Consid
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