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Reil [10]
3 years ago
10

I need to answer three questions about y = x^(-2) on the interval [-4, -1/2]. I have attached my questions and my answers as scr

eenshots:
Part 1 - ANSWER C
Part 2 - ANSWER A
Part 3 - ANSWER B

I am referring to each part in the order that I uploaded the screenshots. You will see that my answers are highlighted in blue. I like second opinions!

Mathematics
1 answer:
77julia77 [94]3 years ago
5 0
I wasn't sure how to order things, so I decided to write directly on top of your screenshots. Have a look at the three attachments. Two of which are correct. There is one incorrect answer though you're fairly close. 

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What's 17x=18(1/3) ?
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<span>17x=18(1/3)
17x = 6
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Use the relationships between the angles to find the value of x (please answer)
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Step-by-step explanation:

x + 15 \degree = 102 \degree \:  \\ (corresponding \:  \angle s) \\  \therefore \: x =  102 \degree -  15 \degree  \\  \huge \red { \boxed{\therefore \: x =  87 \degree}}

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Five to the 25th power equals what
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2 years ago
Given f(x)=e^-x^3 find the vertical and horizontal asymptotes
Aleonysh [2.5K]

Given:

f\mleft(x\mright)=e^{-x^3}

To find the vertical and horizontal asymptotes:

The line x=L is a vertical asymptote of the function f(x) if the limit of the function at this point is infinite.

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Thus, the function f(x) doesn't have a vertical asymptote.

The line y=L is a vertical asymptote of the function f(x) if the limit of the function (either left or right side) at this point is finite.

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Thus, y = 0 is the horizontal asymptote for the given function.

5 0
11 months ago
The number of pizzas ordered on friday evenings between 5:30 and 6:30 at a pizza delivery location for the last 10 weeks is show
Ainat [17]
For a smoothing constant of 0.2

Time period – 1 2 3 4 5 6 7 8 9 10

Actual value – 46 55 39 42 63 54 55 61 52

Forecast – 58 55.6 55.48 52.18 50.15 52.72 52.97 53.38 54.90

Forecast error - -12 -.6 -16.48 – 10.12 12.85 1.28 2.03 7.62 -2.9
The mean square error is 84.12

The mean forecast for period 11 is 54.38
For a smoothing constant of 0.8

Time period – 1 2 3 4 5 6 7 8 9 10

Actual value – 46 55 39 42 63 54 55 61 52

Forecast – 58 48.40 53.68 41.94 41.99 58.80 54.96 54.99 59.80

Forecast error - -12 6.60 -14.68 0.06 21.01 -4.80 0.04 6.01 -7.80The mean square error is 107.17

The mean forecast for period 11 is 53.56


Based on the MSE, smoothing constant of .2 offers a better model since the mean forecast is much better compared to the 53.56 of the smoothing constant of 0.8.
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3 years ago
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