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rosijanka [135]
3 years ago
12

Solve 4 [5 (12+3)-2]-7 PLSSS HELP MEEE

Mathematics
1 answer:
viva [34]3 years ago
3 0

Answer:

285

Step-by-step explanation:

4(5x15-2)-7

4(75-2)-7

4x73-7

292-7

285

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What are three consecutive integers whose product is 480 more than their sum?
Natali [406]

Answer:

The three numbers are 7 8 and 9

Step-by-step explanation:

Givens

  • Let the first number be n - 1
  • Let the second number be n
  • Let the third number = n + 1

Equation

(n - 1)(n)(n + 1) - (n-1 + n + n+1) = 480

Solution

Multiply (n - 1) and (n + 1) = (n - 1)*(n + 1) = n^2 - 1

Multiply the second integer by the result of the first and third: n (n^2 - 1)

Add the three integers together: (x - 1) + (n - 1) + n = 3n  Combine these 2 steps

n(n^2 - 1) - 3n = 480 Remove the brackets

n^3 - n - 3n = 480

n^3 - 4n = 480

n^3 - 4n - 480 = 0

Graph

The graph shows that the intercept point is n =8. This is the only way I can see to solve this cubic. There are no other real roots.

Answer

n - 1 = 7

n = 8

n + 1 = 9

Check

Product 7*8*9 = 504

Sum = 7 + 8 + 9 = 24

504 - 24 = 480 Which checks.

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Z2 + (2j – 3)z + (5 – j) = 0
BARSIC [14]

Answer:

2z+(2j-3)z+(5-j)=0

2z +2jz-3z+5-j=0

2z-3z+2jz-j=-5

-z+2jz-j=-5

2jz-z-j=-5

I hope this help

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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