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lesya692 [45]
3 years ago
12

What is the sum (6x^3 – 5x^2 + 3x – 5) + (8x^4 + 3x^3 + 5x^2 + x + 4)?

Mathematics
2 answers:
nadezda [96]3 years ago
7 0

Answer:

8x^4+9x^3+4x-1

Step-by-step explanation:

(6x^3 – 5x^2 + 3x – 5) + (8x^4 + 3x^3 + 5x^2 + x + 4)

Group like terms

8x^4+6x^3+3x^3-5x^2+5x^2+3x+x-5+4

8x^4+9x^3+4x-1

TiliK225 [7]3 years ago
6 0

Answer:

\left(6x^3-5x^2+3x-5\right)+\left(8x^4+3x^3+5x^2+x+4\right)

Combine like terms

=8x^4+6x^3+3x^3-5x^2+5x^2+3x+x-5+4

Add: -5x²+5x²=0

=8x^4+6x^3+3x^3+3x+x-5+4

Now add 6x³+3x³=9x³

=8x^4+9x^3+3x+x-5+4

Add like terms 3x +x=4x

=8x^4+9x^3+4x-5+4

Now add -5+4=-1

=8x^4+9x^3+4x-1

<u>OAmalOHopeO</u>

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3 years ago
I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

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

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

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so either
-10a = 0 or (-2x+1)^4 = 0
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Bonjour je ne comprend pas cette exercice; si quelqu'un peut m'aider je suis preneuse.
gayaneshka [121]

Answer:

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Tu connais donc un angle et le côté opposé à l'angle et tu cherche le côté adjacent.

On sait que tan(â) = \frac{Cote oppose}{Cote adjacent} donc :

tan(40)= \frac{1,27}{x}

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j'espère avoir été clair.

<h2 />
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