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Alona [7]
3 years ago
5

How to solve vatiable equations 5x-7=7x-17

Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0
Let's get the same terms on the same side.
I am going to remove 5x from both sides of the equation by subtracting 5x

5x - 5x -7 = 7x -5x - 17
           -7 = 2x - 17   now lets remove the 17 from both sides
       -7 + 17 = 2x     (why do we add 17?  Think of 2x -17 as 2x +  -17.  Tor remove a negative number we add it.
Now we finally have     10=2x and if we divide both sides by 2

<u>10</u>  = <u>2x</u><u>
</u>2       2    we get that x = 5
<u />Let's check by substituting 5 in the original equation:
5(5) - 7 = 7(5) - 17
25 - 7 = 35 - 17
18      =       18
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Evaluate if K=3 <br> K^2 + 5= <br> I suck at math please help!!:)
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Answer:

14

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Step-by-step explanation:

<u>Step 1: Define</u>

K² + 5

K = 3

<u>Step 2: Evaluate</u>

  1. Substitute in <em>K</em>:                    3² + 5
  2. Exponents:                           9 + 5
  3. Add:                                      14
3 0
3 years ago
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Write the algebraic expression that matches each graph:
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8 0
4 years ago
Help help help ASAP please quiz math
Slav-nsk [51]

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6x + 18

Step-by-step explanation:

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5 0
2 years ago
1. A certain manufacturing process to build phone batteries results in the production of defective batteries at a rate of 5% (0.
lesya [120]

Answer:

0.5 ; 0.475 ; 0.689 ; 0.4013

Step-by-step explanation:

Given that:

Rate of production of defective batteries p = 0.05

Number of batteries produced (n) = 10

The expected number of defective batteries = mean = n * p = 10 * 0.05 = 0.5 batteries

Variance of defective batteries :

Var(X) = n * p * q ; q = 1 - p

Hence,

Var(X) = 10 * 0.05 * 0.95 = 0.475

Standard deviation (X) = sqrt(variance) = sqrt(0.475) = 0.689

Probability that atleast 1 battery is defective :

Using the binomial probability function

P(x ≥ 1) = 1 - p(x = 0)

= 1 - q^n

= 1 - 0.95^10

= 1 - 0.59873693923837890625

= 0.40126306076162109375

= 0.4013

7 0
3 years ago
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