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ale4655 [162]
3 years ago
10

Find two consecutive odd integers such that their product is 111 more than 3 times their sum​

Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

The numbers are -9 and -7 or 13 and 15

Step-by-step explanation:

Let

x and x+2 ----> two consecutive odd integers

we know that

x(x+2)=3[x+x+2]+111  

Solve for x

x(x+2)=3[x+x+2]+111\\ \\x^{2}+2x=6x+6+111\\ \\x^{2}-4x-117=0

Solve the quadratic equation by graphing

The solution is x=-9, x=13

see the attached figure

<em>First solution</em>

x=-9

x+2=-9+2=-7

The numbers are -9 and -7

<em>Second solution</em>

x=13

x+2=13+2=15

The numbers are 13 and 15

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Tju [1.3M]
None of your options, not via complete the square nor quadratic the formula.

Solve for x over the real numbers:
x^2 + 8 x + 9 = 0

x = (-8 ± sqrt(8^2 - 4×9))/2 = (-8 ± sqrt(64 - 36))/2 = (-8 ± sqrt(28))/2:
x = (-8 + sqrt(28))/2 or x = (-8 - sqrt(28))/2

sqrt(28) = sqrt(4×7) = sqrt(2^2×7) = 2sqrt(7):
x = (2 sqrt(7) - 8)/2 or x = (-2 sqrt(7) - 8)/2

Factor 2 from -8 + 2 sqrt(7) giving 2 (sqrt(7) - 4):
x = 1/22 (sqrt(7) - 4) or x = (-2 sqrt(7) - 8)/2

(2 (sqrt(7) - 4))/2 = sqrt(7) - 4:
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Factor 2 from -8 - 2 sqrt(7) giving 2 (-sqrt(7) - 4):
x = sqrt(7) - 4 or x = 1/22 (-sqrt(7) - 4)

(2 (-sqrt(7) - 4))/2 = -sqrt(7) - 4:

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Solve for x:
x^2 + 8 x + 9 = 0

Subtract 9 from both sides:
x^2 + 8 x = -9

Add 16 to both sides:
x^2 + 8 x + 16 = 7

Write the left hand side as a square:
(x + 4)^2 = 7

Take the square root of both sides:
x + 4 = sqrt(7) or x + 4 = -sqrt(7)

Subtract 4 from both sides:
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Subtract 4 from both sides:
Answer:  x = sqrt(7) - 4 or x = -4 - sqrt(7)
4 0
3 years ago
I will give multiple brainlist to people who help me with these slides!
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Answer:

A)6

B)ST

Step-by-step explanation:

A if u turn it,  15 is equal to the 10

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x=6

B) Turn it so #2 is facing up.

ST is the same.

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4 0
2 years ago
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SashulF [63]

2(x-5)=20

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Answer

x = 15 is the solution

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0.06 is 10 times as much as 0.6
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xxTIMURxx [149]

Answer:

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