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vlabodo [156]
3 years ago
15

Simplify (8j3 + 3j2 + 7) – (3j3 – 7j2 – 7j – 12)

Mathematics
2 answers:
ehidna [41]3 years ago
4 0
8j³+3j²+7-(3j³-7j²-7j-12)
8j³+3j²+7-3j³+7j²+7j+12
8j³-3j³+3j²+7j²+7j+7+12
5j³+10j²+7j+19
KengaRu [80]3 years ago
4 0

Answer:

5j^3 + 10j^2 +7j+19

Step-by-step explanation:

(8j^3 + 3j^2 + 7) - (3j^3 - 7j^2 - 7j - 12)

To simplify it we need to remove the parenthesis

WE multiply the negative sign inside the parenthesis'

8j^3 + 3j^2 + 7- 3j^3 + 7j^2 + 7j + 12

Now we combine like terms

8j^3 - 3j^3 =5j^3

3j^2 +7j^2=10j^2

7+12=19

Expression becomes

5j^3 + 10j^2 +7j+ 19

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One positive integer is 1 less than twice another. The sum of their squares is 106. Find the integers.
neonofarm [45]
They give us 2 pieces to the puzzle.  Both are positive numbers...x and y.
1.) 1 number is 1 less than twice another number. (x = 2y -1)...and
2.) the sum of their squares is 106.  (x^2 + y^2 = 106).

 substitute the value for x into the second equation.

(2y-1)^2 + y^2 = 106
(2y-1) (2y-1) + y^2 = 106  (use distributive property)
4y^2 - 2y - 2y + 1 + y^2 = 106  (subtract 106 from both sides)
4y^2 - 2y - 2y + 1 + y^2 - 106 = 106 - 106 (combine like terms)
5y^2 - 4y - 105 = 0  (factor)
(y-5) (5y-21) = 0  (set to 0)
y - 5 = 0
y = 5 

substitute the 5 into the equation for y   (x = 2(5) - 1)
           x = 9   if we square 9, we get 81.
subtracted from 106 we have 25...the square root of 25 is 5.

our answers are 5 and 9.
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Sliva [168]

Answer:

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5x³ - 21 + 7

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