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2 years ago

Find four consecutive integers such that the sum of the second and fourth is 132

Mathematics

1 answer:

Ulleksa [173]2 years ago

7 0

Remark

As long as the representation shows the numbers come one after another, you can get x to be anything.

Givens

x Can be the second number

x - 1 Can be the first number

x + 1 Can be the third number

x + 2 Can be the fourth number

Equation

x + x + 2= 132

Solution

2x + 2 = 132

2x = 130

x = 65

So the 4 numbers are 64 65 66 67

Check

The second and fourth numbers added together = 65 + 67 = 132

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3 years ago

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You know 4−3i is the complex conjugate of 4+3i. If you multiply the denominator of −2+5i4+3i by 4−3i, what will you get? Enter y

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Answer:

Step-by-step explanation:

Given

(-2 + 5i)/(4 + 3i)

Required

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The denominator of the fraction above is 4 + 3i

When multiplied by 4 - 3i, the result is as follows;.

Result = (4 + 3i)(4 - 3i)

Expand

Result = 4(4 - 3i) + 3i(4 - 3i)

Open both brackets (take it one at a time)

Result = 4 * 4 - 4 * 3i + 3i(4 - 3i)

Result = 16 - 12i + 3i(4 - 3i)

Open the second bracket

Result = 16 - 12i + 12i - 3i * 3i

Result = 16 - 12i + 12i - 9i²

In complex numbers, i = √-1

i² = (√-1)²

i² = √-1 * √-1

i² = -1

The expression

Result = 16 - 12i + 12i - 9i²

Becomes

Result = 16 - 12i + 12i - 9(-1)

Result = 16 - 12i + 12i + 9

Collect like terms

Result = 16 + 9 + 12i - 12i

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