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Reil [10]
3 years ago
7

We use the properties of equality to rewrite equations and solve equations for a variable. These properties allow us to maintain

a balance between the two sides of the equation.
Why is it important to keep the two sides of an equation balanced when solving? What other properties do we use to rewrite expressions and equations?
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
6 0
Help ASAP only right answers only no spam don’t answer if you don’t know

















Important usd
You might be interested in
Branilest and 30 point please answer
Lynna [10]

Answer:

<em>Addition Property of Equality</em>

Step-by-step explanation:

<u>System of Equations</u>

When two equations are given and two variables are unknown in both equations, then we can solve the system in several ways.

One method consists in adding both equations term by term and eliminate one of the variables. That property is called the addition of equalities. Let's consider the equations given in the problem:

5x+4y=-2\\2x-4y=6

If we add both equations, we have

5x+4y+2x-4y=-2+6

Simplifying

7x=4

By adding both equations we managed to eliminate the variable y and could easily find the value of x

x=4/7

Answer: Addition Property of Equality

7 0
3 years ago
Please help meeeee :,(
Crank

Answer:

I pretty sure the answer is <u>A</u><u>S</u><u>A</u>, I hope it's right!

6 0
3 years ago
It takes Aaron 5 minutes to squeeze the juice from 3 grapefruits. At this rate, how many minutes will it take him to squeeze the
aliya0001 [1]
20 minutes because 12/3=4 and 4x5=20
3 0
3 years ago
A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
Fed [463]

Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

The initial conditions are : a_{0} = a_{1} = 0

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

7 0
3 years ago
What is 45% of 8.5 ?
satela [25.4K]

Answer:

3.825

Step-by-step explanation:

First, you have to move the decimal two over on 45%. If you move the decimal over two spaces to the right, it would be 0.45. In math, of means to multiply, so then you multiply 0.45 and 8.5 to get your answer of 3.825.

7 0
3 years ago
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