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1 year ago

OMG PLS HELP ME!!!!!! I NEED THIS ASAP!!! 100 PTS!!!!!!! (its in the pic)

Mathematics

1 answer:

Rainbow [258]1 year ago

5 0

By algebra properties we find the following relationships between each pair of algebraic expressions:

First equation: Case 4
Second equation: Case 1
Third equation: Case 2
Fourth equation: Case 5
Fifth equation: Case 3

<h3>How to determine pairs of equivalent equations</h3>

In this we must determine the equivalent algebraic expression related to given expressions, this can be done by applying algebra properties on equations from the second column until equivalent expression is found. Now we proceed to find for each case:

First equation

(7 - 2 · x) + (3 · x - 11)

(7 - 11) + (- 2 · x + 3 · x)

- 4 + (- 2 + 3) · x

- 4 + (1) · x

- 4 + (5 - 4) · x

- 4 - 4 · x + 5 · x

- 4 · (x + 1) + 5 · x → Case 4

Second equation

- 7 + 6 · x - 4 · x + 3

(6 · x - 4 · x) + (- 7 + 3)

(6 - 4) · x - 4

2 · x - 4

2 · (x - 2) → Case 1

Third equation

9 · x - 2 · (3 · x - 3)

9 · x - 6 · x + 6

3 · x + 6

(2 + 1) · x + (14 - 8)

[1 - (- 2)] · x + (14 - 8)

(x + 14) - (8 - 2 · x) → Case 2

Fourth equation

- 3 · x + 6 + 4 · x

x + 6

(5 - 4) · x + (7 - 1)

(7 + 5 · x) + (- 4 · x - 1) → Case 5

Fifth equation

- 2 · x + 9 + 5 · x + 6

3 · x + 15

3 · (x + 5) → Case 3

To learn more on algebraic equations: brainly.com/question/24875240

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x^2=0
x=0
zeroes are at (0,0)

for p(x)/q(x)

when degree of p(x)<q(x), HA=0

when degree of p(x)=q(x), HA= leading coef of p(x) divided by leading coef of q(x)

when degree of p(x)>q(x) you probably have a slant assymtote

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leading coefs

1/1=1

HA is y=1

does it cross?

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graph the point (0,0)

draw the lines

y=1

x=-2

x=2

reemmber, plus, minus, plus

so from left, it goes from above the HA right up to the VA of x=-2

then goes upside down U shape in between VA's going through (0,0)

then from top of VA x=2 down to y=1 then gets closer to y=1 but never touches

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