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2 years ago

The sum of 139 and the product of a number raised to the third power and 17. In symbols, the phrase is written as.

Mathematics

1 answer:

qwelly [4]2 years ago

7 0

Answer:

Step-by-step explanation:

The word "sum" means the amount resulting from addition in an equation. This means that we are adding all of these numbers together.

The word "product" means the amount resulting from multiplication in an equation, this means we need to find the product of a number multiplied by itself three times. Since we don't know what number it is, we can use the variable 'x'.

So, you would be adding 139, the product of x³, and 17, which would be written as y = 139 + x³ + 17.

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4 years ago

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3 years ago

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Novay_Z [31]

Answer:

$\frac{\sqrt{3} - 1}{2\sqrt{2}}$

$\frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

$- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Step-by-step explanation:

Given $\frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6}$

(A) $sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6})$

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2}$

$$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

(B) Cos(A + B) = CosAcosB - SinASinB

$cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}})$

$\implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2}$

$\implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6})$

$$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

From the above obtained values this can be calculated and the value is $- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .

3 0

3 years ago