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3 years ago

Solving systems by substitutions

1 answer:

8 0

Let's take two different equations.

4x+2y=8
7x+3y=-10
(I just came up with these on the spot, so the answers might be strange but i should be able to explain nevertheless)
First, let's solve the first equation for "y".
4x+2y=8
-4x -4x
2y=8-4x
/2
y=4-2x
Now that we have "y", let's plug it into the second equation.
7x+3(4-2x)=-10
7x+12-6x=-10
x+12=-10
-12 -12
x=-22
Alright, now we have "x" definitively, let's plug it in to find "y".
4(-22)+2y=8
-88+2y=8
+88 +88
2y=96
/2
y=48
So from this example you can see. You solve one variable in terms of another (solve y as being 4-2x), plug it into the second equation to find one variable's value, then plug that variable's value back in again later to get the first variable's value.

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Jim bought the game system on sale for $195. Kaylen bought the same system at full price for $250. How much was the sale for whe

castortr0y [4]

Answer:

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I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

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3 years ago