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Nady [450]
3 years ago
8

If f(x)=x+4 and g(x)=x^3 what is (g*f)(-3)

Mathematics
1 answer:
Molodets [167]3 years ago
4 0

Step-by-step explanation:

g*f(x)=g(x+4)=(x+4)³

g*f(-3)=(-3+4)³

= 1³=1

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Consider the system of equations:
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\large{\underline{\underline{\pmb{\sf {\color {blue}{Solution:}}}}}}

Since, we know that in Elimination method we have to first the value of "x" or either "y". For that we have to multiply a number which makes the both equation's "x" or "y" equal so that we cut cancel it and find the solution.

Here's an example:

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Equation 1: - 2x + 3y = 9

Equation 2: - 8x - 7y = 10

We can see that in "Equation 1" the first number is "- 2x" and in "Equation 2" the first number is "-8x". So, what we do in Elimination method is that we have to make the first number of both the equations equal or same.

Eg:

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So, I'm taking the number "4" to multiply it with equation 1, which gives us the result,

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Let a and b be two positive numbers. If 2a + 3b=6 then the maximum product of these a and b is:
VikaD [51]

Answer:

a= 3/2

b= 1

a*b= 3/2

Step-by-step explanation:

For getting this problem we need to think in terms of only one unknown. Lets start by looking at our equation:

2a + 3b=6

No, we need to see what happens to a*b. Lets get the value of one of these unknowns in terms of the other, this is, a in terms of b or b in terms of a. Here, I will get a in terms of b (if you like, do the other way and compare):

2a + 3b=6

Subtract 3b in both sides:

2a = 6 - 3b

Now divide both sides by 2 to get a alone:

a = (6 - 3b)/2

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Now lets try to multiply a by b, using this a we found above:

a*b = (3 - 3/2 b)*b

Using distributive:

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So, we have an expression for a*b that depends only on b. Notice that this expression is a parabola with negative coefficient on the square term, what makes it a negative parabola that MUST have a maximum value. So, we have an expression for a*b that can be maximized, so we can find the maximum of a*b by the derivative of the expression. Lets derive in b:

(3b - 3/2 b^2)' = (3b)' - (3/2 b^2)' = 3 - 3b

So, the derivative equal to 0 gives us the maximum:

3 - 3b = 0

Suming 3b in both sides:

3 = 3b

Dividing by 3:

1 = b

So, we maximize our expression a*b when b is equal to 1. Now, we can replace it on a = 3 - 3/2 b to find b:

a = 3 - 3/2(1) = 3 - 3/2 = 3/2

Thus, a is equal to 3/2 and the product a*b is maximized when:

a = 3/2

b = 1

And the product is a*b= (3/2) * 1 = 3/2

5 0
3 years ago
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