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Umnica [9.8K]
3 years ago
8

Please help me out with this!!

Mathematics
1 answer:
finlep [7]3 years ago
6 0

Answer:

see attached

Step-by-step explanation:

<h3>1.</h3>

Solid dots go at all the end points except the one at (3, 6), which gets an open dot signifying the function is not defined for that point.

For the first portion of the graph, the square root function is shifted left one unit and scaled vertically by a factor of 3.

The second portion of the graph is a line with a slope of -1. It would have a y-intercept of 5 if it were defined there. It has an x-intercept of 5.

<h3>2.</h3>

The y-intercept is found by setting x=0 and solving for y.

... y = log(2·0+1) -1 = log(1) -1 = 0 -1 = -1

The x-intercept is found by setting y=0 and solving for x.

... 0 = log(2x +1) -1

... 1 = log(2x +1) . . . add 1

... 10 = 2x +1 . . . . . . take the antilog

... x = (10 -1)/2 = 4.5 . . . . . subtract 1, divide by the x coefficient

The intercepts are (0, -1) and (4.5, 0).

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Can someone please explain how to do these?
Pani-rosa [81]

Answer:

<u>First question answer:</u> The limit is 69

<u>Second question answer:</u> The limit is 5


Step-by-step explanation:

For the first limit, plug in x=8 in the expression (9x-3), that's the answer for linear equations and limits.

So we have:

9x-3\\9(8)-3\\72-3\\69

The answer is 69


For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value x=1 into the simplified expression to get the correct answer. Shown below:

\frac{x^2+8x-9}{x^2-1}\\\frac{(x+9)(x-1)}{(x-1)(x+1)}\\\frac{x+9}{x+1}

<em>Now putting 1 in x gives us the limit:</em>

\frac{x+9}{x+1}\\\frac{1+9}{1+1}=\frac{10}{2}=5

So the answer is 5

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What is the domain and range for the following function and its inverse?
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Step-by-step explanation:

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4 years ago
Please help! x^3 +y^3=9
Crank

Step-by-step explanation:

Taking the derivative of both sides with respect to x, we get

\dfrac{d}{dx}(x^3 + y^3) = \dfrac{d}{dx}(3)

\Rightarrow 3x^2 + 3y^2\dfrac{dy}{dx} = 0

Solving for \frac{dy}{dx}, we get

\dfrac{dy}{dx} = -\dfrac{3x^2}{3y^2} = -\dfrac{x^2}{y^2}

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