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3 years ago

Subtract the cube root of the product of x and 3y from the square of the sum of a and b.

Mathematics

1 answer:

rosijanka [135]3 years ago

7 0

Answer:

a^2 + b^2 + 2ab - (3xy)^1/3

Step-by-step explanation:

Here we want to make a subtraction

Cube root of the product of x and 3y

x * 3y = 3xy

Cube root of this;

(3xy)^1/3

The sum of a and b is (a + b)

Square of this sum;

(a + b)^2 = a^2 + 2ab + b^2

Now, subtract the cube root

we have;

a^2 + b^2 + 2ab - (3xy)^1/3

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Anuta_ua [19.1K]

Answer:

$\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}$

Step-by-step explanation:

Given

$\sin(105^o)$

Required

Solve

Using sine rule, we have:

$\sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)$

This gives:

$\sin(105^o) = \sin(60 + 45)$

So, we have:

$\sin(60 + 45) = \sin(60)\cos(45) + \sin(45)\cos(60)$

In radical forms, we have:

$\sin(60 + 45) = \frac{\sqrt 3}{2} * \frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} * \frac{1}{2}$

$\sin(60 + 45) = \frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}$

Take LCM

$\sin(60 + 45) = \frac{\sqrt 6 + \sqrt 2}{4}$

Rewrite as:

$\sin(60 + 45) = \frac{\sqrt 2 + \sqrt 6}{4}$

Hence:

$\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}$

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3 years ago

Last year, there were 148 pies baked for the bake sale. This year, there were c pies baked. Using c, write an expression for the

solniwko [45]

$\huge{\boxed{c+148}}$

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Answer:

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Step-by-step explanation:

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