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Scrat [10]
3 years ago
9

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-4, -2), B(-2, -2), C(-2,

7)
Mathematics
1 answer:
Readme [11.4K]3 years ago
7 0
What’s your question
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Someone please help me thank you
alexandr1967 [171]
<h3>Answer: Slope = -3/4</h3>

slope = rise/run

rise = -3 ... move down 3 units

run = 4 .... move to the right 4 units

============================================

Work Shown:

m = slope

m = rise/run

m = (change in y)/(change in x)

m = (y2-y1)/(x2-x1)

m = (-17-(-2))/(24-4)

m = (-17+2)/(24-4)

m = -15/20

m = -3/4 is the slope

----------------

It basically means "move down 3 units, and move to the right 4 units"

So, slope = rise/run = -3/4

rise = -3 = move down 3 units

run = 4 = move to the right 4

You could express the slope as a decimal, and that would be -3/4 = -0.75, but you lose the rise and run values. So I recommend you keep the fraction.

Another problem with decimals is that sometimes we could have approximations that aren't exact. For example, if the slope was 2/3, then that approximates to 0.667 which may cause rounding errors down the line. Though I should point out that decimals do have their advantages in some cases.

5 0
3 years ago
Exponents of the multiple 3
nlexa [21]

I believe you are referring to exponents that have a multiple of three. If so any number * three is a valid exponent in your case.

4 0
3 years ago
Can someone help me with these questions please and thank you.
CaHeK987 [17]

Answer: 1st is 16 2nd is 48 the 3rd one is 8

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Expand:<br><img src="https://tex.z-dn.net/?f=f%28z%29%20%3D%20%20%5Cfrac%7B1%7D%7Bz%28z%20-%202%29%7D%20" id="TexFormula1" title
Monica [59]

Expand f(z) into partial fractions:

\dfrac1{z(z-2)} = \dfrac12 \left(\dfrac1{z-2} - \dfrac1z\right)

Recall that for |z| < 1, we have the power series

\displaystyle \frac1{1-z} = \sum_{n=0}^\infty z^n

Then for |z| > 2, or |1/(z/2)| = |2/z| < 1, we have

\displaystyle \frac1{z-2} = \frac1z \frac1{1 - \frac2z} = \frac1z \sum_{n=0}^\infty \left(\frac 2z\right)^n = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}

So the series expansion of f(z) for |z| > 2 is

\displaystyle f(z) = \frac12 \left(\sum_{n=0}^\infty \frac{2^n}{z^{n+1}} - \frac1z\right)

\displaystyle f(z) = \frac12 \sum_{n=1}^\infty \frac{2^n}{z^{n+1}}

\displaystyle f(z) = \sum_{n=1}^\infty \frac{2^{n-1}}{z^{n+1}}

\displaystyle \boxed{f(z) = \frac14 \sum_{n=2}^\infty \frac{2^n}{z^n} = \frac1{z^2} + \frac2{z^3} + \frac4{z^4} + \cdots}

6 0
2 years ago
The value of square root -9 is not -3 because ___
LUCKY_DIMON [66]

Answer:

you cannot have a negative root. It errors out if you put the negative in the root.

Hope that helps feel free to ask more questions


Brainliest??

Step-by-step explanation:


4 0
3 years ago
Read 2 more answers
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