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3 years ago

the sum of three consecutive integers is 78. What is the value of the smallest of the three integers?

Mathematics

1 answer:

CaHeK987 [17]3 years ago

8 0

Answer:

Step-by-step explanation:

Consecutive integers just means like one after another.

This means that:

First number = x

Second number = x + 1

Third number = x + 2

They've told us that the sum of 3 consecutive integers is 78.

First number + second number + third number = 78

Sub the equations we wrote into the formula above and we get:

x + x + 1 + x + 2 = 78

Gather the like terms.

3x + 3 = 78

Subtract 3 from both sides.

3x = 78-3 = 75

Divide both sides by 3 to solve for x.

x = 75/3 = 25

If you look at the very beginning we said that the first number = x.

Therefore it is 25.

Second number = x + 1 = 26

Third number = x + 2 = 27

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Write the sentences as proportion. Twelve hundredths is to three as twenty-four hundredths is to eight

ValentinkaMS [17]

Answer:

see the explanation

Step-by-step explanation:

Remember that

A proportion is the ratio (quotient) between two numbers

Twelve hundredths to three is ----> $\frac{0.12}{3}$

Twenty-four hundredths to eight ----> $\frac{0.24}{8}$

therefore

Twelve hundredths is to three as twenty-four hundredths is to eight

$\frac{0.12}{3}=\frac{0.24}{8}$

Note The proportion is not true

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2 years ago

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Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

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2 years ago

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kvv77 [185]

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Step-by-step explanation:

You take the first part, "perpendicular lines intersect", and add the "if" to it. Then, you take the second part, "to form right angles", and add the "then" in front of it to get your answer, "If perpendicular lines intersect, then they form right angles."

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Answer:

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Step-by-step explanation:

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