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Julli [10]
3 years ago
7

F(x) = | 3x +4 |+2 g(x)=4 Find (f +g)(x)

Mathematics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

|3x + 4 | + 6

Step-by-step explanation:

(f+g)(x)= f(x) + g(x)

= |3x+4| + 2 + 4

= | 3x + 4 | + 6

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Peter has 3200 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed a
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Answer:

A = 640000\,yd^{2}

Step-by-step explanation:

Expression for the rectangular area and perimeter are, respectively:

A (x,y) = x\cdot y

3200\,yd = 2\cdot (x+y)

After some algebraic manipulation, area expression can be reduce to an one-variable form:

y = 1600 -x

A (x) = x\cdot (1600-x)

The first derivative of the previous equation is:

\frac{dA}{dx}= 1600-2\cdot x

Let the expression be equalized to zero:

1600-2\cdot x=0

x = 800

The second derivative is:

\frac{d^{2}A}{dx^{2}} = -2

According to the Second Derivative Test, the critical value found in previous steps is a maximum. Then:

y = 800

The maximum area is:

A = (800\,yd)\cdot (800\,yd)

A = 640000\,yd^{2}

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3 years ago
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Complete the remainder of the table for the given function rule: y= x/4 + 7
babymother [125]

Let's replace x with -4 and simplify

y = (x/4) + 7

y = (-4/4) + 7

y = -1 + 7

y = 6

This means x = -4 and y = 6 pair up together.

So 6 goes in the box under the -4

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

Repeat those steps for x = 0

y = (x/4) + 7

y = (0/4) + 7

y = 0 + 7

y = 7

So 7 goes in the box under the 0

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Let's repeat those steps for x = 4

y = (x/4) + 7

y = (4/4) + 7

y = 1 + 7

y = 8

So 8 goes in the box under the 4

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Lastly, for x = 8, we get

y = (x/4) + 7

y = (8/4) + 7

y = 2 + 7

y = 9

9 goes in the box under the 8

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The y outputs we got were: 6, 7, 8, 9

Notice how each time y increases by 1, x goes up by 4.

This means slope = rise/run = 1/4

We can think of x/4 as (1/4)x to help see that the slope is 1/4.

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2 years ago
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Step-by-step explanation:

Put 1 on the far left and 4 on the far right.

Put. 2 the top and 3 at the bottom.

Put 5 in the centre.

There are 2 ways you can find to to make all the numbers add up to 5.

The number in the centre is the common anwser to the sums.

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