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3 years ago

Find formula for g(x) if it is known that g(x-3)=2x-6

Mathematics

1 answer:

kati45 [8]3 years ago

7 0

9514 1404 393

Answer:

g(x) = 2x

Step-by-step explanation:

Replace x in the given function with x+3.

g((x+3) -3) = 2(x+3) -6

g(x) = 2x +6 -6

g(x) = 2x

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Determine the intercepts of the line that correspond to the following table of values.

Dahasolnce [82]

We know that
to find the equation of a line i need two points

let
point A-----> (-21,10)
point B-----> (-28,14)

step 1
find the slope m
m=(y2-y1)/(x2-x1)-----> m=(14-10)/(-28+21)----> m=-4/7

step 2
find the equation of a line with
m=-4/7 and the point A (-21,10)
y-y1=m*(x-x1)------> y-10=(-4/7)*(x+21)---> y=(-4/7)x-12+10
y=(-4/7)x-2

step 3
find x intercept
for y=0
y=(-4/7)x-2------> 0=(-4/7)x-2-----> 4/7x=-2----> x=-14/4----> x=-3.5

the x intercept is (-3.5,0)

step 4
find the y intercept
for x=0
y=(-4/7)x-2--------> y=-2
the y intercept is (0,-2)

the answers are
the x intercept is (-3.5,0)
the y intercept is (0,-2)

see the attached figure

7 0

2 years ago

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

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1 year ago

Jason rented 4 movies. Jason plans to watch all 4 movies back to back. In how many different orders can Jason watch the movies?

Aloiza [94]

Answer:

Step-by-step explanation:

i think it’s this because 4(4) is 16?

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3 years ago

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yKpoI14uk [10]

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A researcher administers a treatment to a sample of n = 25 participants and uses a hypothesis test to evaluate the effect of the

RideAnS [48]

Answer: option B

Step-by-step explanation: at alpha level of 5% for a two tailed test, the critical value is +1.96 and -1.96

At alpha level of 1% for a two tailed test, the critical value is +2.576 and -2.576

Relating z = 2.77 with these critical values, we can see that at 5% level of significance, z = 2.77 is greater than +1.96 which falls in the rejection region.

Also at 1% level of significance, we can see that z = 2.77 is greater than 2.576 which also falls in the rejection region.

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3 years ago

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