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Tom [10]
2 years ago
13

Drag the correct classification for each graph into the boxes to complete the table.

Mathematics
2 answers:
Andrei [34K]2 years ago
6 0

Answer:

Odd, neither, neither

Step-by-step explanation:

A function is even if f(x) = f(-x).  That means that it passes through the origin and is symmetrical about the y-axis.

A function is odd if f(x) = -f(-x).  That means that it passes through the origin and is symmetrical about the origin.

The first graph, the line, passes through the origin and is symmetrical about the origin.  So it is odd.

The second graph does not pass through the origin, nor is it symmetrical.  So it is neither odd nor even.

The third graph does not pass through the origin, nor is it symmetrical about the y-axis.  So it is neither odd nor even.

tester [92]2 years ago
5 0

neither even or odd, odd, even

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salantis [7]
To solve this system of linear equations we need a little trick called elimination steps are below

we are given the following system of equations
- 4x - 15y = - 17
- x + 5y = - 13
if you look at the x variables in both equations you can see that we can easily eliminate them by multiplying the bottom equation by -4 as so
- 4x - 15y = - 17
- 4( - x + 5y = - 13)
simplify and we get
- 4x - 15y = - 17
4x - 20y = 52
now we can combine like terms on each side with the x's cancelling and get

- 35y = 35
now divide off -35
y = - 1
great now we can go back to our original system and pick a equation and substitute y back in to find x lets use

- x + 5y = 13
so now we substitute y and get
- x + 5( - 1) = 13
- x - 5 = 13
- x = 17
x = - 17

now we put x and y into one coordinate (x,y)
so now our FINAL ANSWER IS
(-17,-1)
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2 years ago
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Show all work to multiply (6+square root -64)(3-square root -16)
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Answer:

50

Step-by-step explanation:

- 8i = √-64

8i + 6

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[8i + 6][-4i + 3] → FOIL

>> -32i² + [24i - 24i] + 18

↑ ↑

-1 0

>>> 32 + 18 = 50

Information on Imaginary Numbers

√-1 = i

-1 = i²

-i = i³

1 = i⁴ [And every other exponent that is a multiple of 4; this cycle then repeats itself every time you go up one number at a time]

If you are ever in need of assistance, do not hesitate to let me know by subscribing to my You-Tube channel [USERNAME: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

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