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otez555 [7]
3 years ago
8

How do you factorise xy zy?

Mathematics
1 answer:
alexandr1967 [171]3 years ago
4 0
Assuming you meant to write: "xy + zy"
The answer is: xy + zy = y(x + z)
The one variable that appear multiple times is: "y". There are no coefficients (other than the implied coefficients of "1"; since "1" multiplied by any number equals that very number).  Since "y" appears twice, we can "factor out" a "y" ; as follows: 
________________
y(x+z).
____________________
Take note of the distributive property of multiplication:
_____________________
a(b+c) = ab + ac ;
_________________
AND: a(b-c) = ab - ac.
____________________
Likewise: y(x+z) = yx +yz 
                        = xy zy
________________________

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in a 3-digit number, the hunderds digit is one more than the ones digit and the tens digit is twice the hundreds digit. If the s
Stels [109]

Answer:362

Step-by-step explanation:

So, the ones digit is 2. Since the hundreds digit is one more, it's 3. And because the tens is twice the hundreds, it's 6. The number is 362

5 0
3 years ago
Who wants a brainist whats the how do you convert ratios
iren2701 [21]

Answer:

  1. To a whole number: Add the numbers in the ratio together
  2. To a fraction: Each number in the ratio would become a numerator in a fraction

I hope this helps!

Credit goes to: calculatorsoup.com

4 0
3 years ago
10 points!!! Please help!!!!!!!
babunello [35]
First find slope
5-10/-1-0= 5

find y-intercept
10=5(0)+B
b=-10

equation: y=5x-10
5 0
4 years ago
Read 2 more answers
Find the standard form of the equation of the parabola with a focus at (0, -9) and a directrix y = 9.
ololo11 [35]

The Standard Form Is

x^2 = 0 (y - 9)

5 0
3 years ago
Read 2 more answers
Simplify (Are all division)<br><br>2+4i<br>3i<br><br>3+2i<br>4+i<br><br>2i^11
Alla [95]

We\ know:\ i=\sqrt{-1}\to i^2=-1

\dfrac{2+4i}{3i}=\dfrac{2+4i}{3i}\cdot\dfrac{3i}{3i}=\dfrac{6i+12i^2}{9i^2}=\dfrac{6i+12(-1)}{9(-1)}\\\\=\dfrac{6i-12}{-9}=\dfrac{2i-4}{-3}=\dfrac{4}{3}-\dfrac{2}{3}i

\dfrac{3+2i}{4+i}=\dfrac{3+2i}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{(3+2i)(4-i)}{4^2-i^2}\\\\=\dfrac{(3)(4)+(3)(-i)+(2i)(4)+(2i)(-i)}{16-(-1)}=\dfrac{12-3i+8i-2i^2}{16+1}\\\\=\dfrac{12+5i-2(-1)}{17}=\dfrac{12+5i+2}{17}=\dfrac{14+5i}{17}=\dfrac{14}{17}+\dfrac{5}{17}i

2i^{11}=2i^{10+1}=2i^{10}i^1=2i^{2\cdot5}i=2i(i^2)^5=2i(-1)^5=2i(-1)=-2i


Used:\ (a+b)(a-b)=a^2-b^2

6 0
3 years ago
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