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valkas [14]
3 years ago
6

Identify the graph of y=x3.

Mathematics
2 answers:
andrew11 [14]3 years ago
7 0

Answer:

EDGE2020

Step-by-step explanation:

graph 2

domain: all real numbers

range: all real numbers

77julia77 [94]3 years ago
6 0

Answer:

2nd Graph

Step-by-step explanation:

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Can someone please help me
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So J is jelly beans and T is trail mix :)
so J= $3.50
T= $1.50

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3 years ago
Calculate the amount of work done if you use a 150N force to push a 50kg box 5m across the kitchen floor.
anzhelika [568]

Answer:

750 J

Step-by-step explanation:

4 0
3 years ago
Roll a number cube. If the number cube comes up odd, you win the same number of points as the number on the cube. If the number
laila [671]

Answer:

–0.5

Step-by-step explanation:

Expected value is found by multiplying the probabilities of each roll by the amount won or lost with each roll.

The odd rolls are 1, 3 and 5.  The probability of rolling a 1 is 1/6, and the value is 1; this gives us 1/6(1) = 1/6.

The probability of rolling a 3 is 1/6, and the value is 3; this gives us 1/6(3) = 3/6.

The probability of rolling a 5 is 1/6, and the value is 5; this gives us 1/6(5) = 5/6.

Rolling any even number gives us the same value.  The probability of rolling an odd number is 3/6, and the value is -4; this gives us 3/6(-4) = -12/6.

Together we have

1/6+3/6+5/6-12/6 = 4/6+5/6-12/6 = 9/6-12/6 = -3/6 = -0.5

5 0
3 years ago
Read 2 more answers
Write a system of linear inequalities represented by the graph?<br> Inequality 1:<br> Inequality 2:
alina1380 [7]

Answer:

y > x - 2

y >= 5x + 1

Step-by-step explanation:

8 0
2 years ago
Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(
Stells [14]

Answer:

f(x) =  {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x  \rightarrow \: x = sec \: y\\  \frac{dx}{dx}  =  \frac{d(sec \: y)}{dx}  \\ 1 = \frac{d(sec \: y)}{dx} \times  \frac{dy}{dy}  \\ 1 = \frac{d(sec \: y)}{dy} \times  \frac{dy}{dx}  \\1 = tan \: y.sec \: y. \frac{dy}{dx}  \\ \frac{dy}{dx} =  \frac{1}{tan \: y.sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ \sqrt{( {sec}^{2}   \: y - 1}) .sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\   \therefore  \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx}  =  \frac{1}{ |5x |  \sqrt{25 {x }^{2}  - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} }

4 0
2 years ago
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