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Wittaler [7]
3 years ago
8

Find four consecutive integers whose sum is 114.

Mathematics
1 answer:
ziro4ka [17]3 years ago
6 0

27, 28, 29 and 30

consecutive integers have a difference of 1 between each one

let n be the first integer then the next three are

n + 1, n + 2 and n + 3

hence n + n + 1 + n + 2 + n + 3 = 114

4n + 6 = 114

subtract 6 from both sides of the equation

4n = 114 - 6 = 108

divide both sides by 4

n = \frac{108}{4} = 27

n = 27, n+ 1 = 28, n + 2 = 29, n + 3 = 30

the consecutive integers are 27, 28, 29 and 30



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Scorpion4ik [409]

Hello! :)

To solve your equation, we will first simplify both sides of the equation

-7 = -1 + x/3

-7 = -1 + 1/3x

-7 = 1/3x - 1

Next, we will flip the equation

1/3x - 1 = -7

Third, we will add 1 to both sides

1/3x - 1 + 1 = -7 + 1

1/3x = -6

Last, we will Multiply both sides by 3

3 * (1/3x) = (3) * (-6)

x = -18 (ANSWER)

That means -18 is you answer

Hope this helped you!

THEDIPER

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1. Kalena was asked to prove ifx(x - 1)(x + 1) = x3 - X represents a polynomial identity. She
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Answer:

C. Kalena made a mistake in Step 3. The justification should state: -x²

+ x²

Step-by-step explanation:

Given the function x(x - 1)(x + 1) = x3 - X

To justify kelena proof

We will need to show if the two equations are equal.

Starting from the RHS with function x³-x

First we will factor out the common factor which is 'x' to have;

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Factorising x²-1 using the difference of two square will give;

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Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;

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Factorising x²-1 using the difference of two square will give;

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Factorising x(x+1) gives x²+x, therefore

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(x²+x)(x-1) = x³-x²+x²-x

The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²

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Step-by-step explanation:

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