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2 years ago

-2/3 + 9/2 what's the answer

Mathematics

2 answers:

Darina [25.2K]2 years ago

4 0

Answer:

The answer would be 23/6

dexar [7]2 years ago

3 0

Not sure I just need points

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Geometry help!!<br><br> I need help on #13, find the values of the variables:

ivanzaharov [21]

T and W would both equal 60

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2 years ago

Read 2 more answers

An arch in the firm of a semicircle was over a 40 inch wide doorway. Find the length of the arch.

Hoochie [10]

Answer:13

Step-by-step explanation:

5 0

2 years ago

12 is 15% of what number?

LenaWriter [7]

The 15% is 1.8

All you have to do is multiply 12 by .15 to get the answer, which is 1.80

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3 years ago

What is the factored form x^3 - 1 ?

amid [387]

Answer: x^3 - 1 = (x - 1)(x^2 + x + 1)

Explanation:

This is a type of factorizing called the sum or difference of 2 cubes:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

The sum of the cubes is factored as:

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

In this case, we have: x^3 - 1 so follow the rule above.

x^3 - 1 = (x - 1)(x^2 + x + 1)

7 0

3 years ago

Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2

Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ .

The sine difference identity is:

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a)$ .

The cosine addition identity is:

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ .

The cosine difference identity is:

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ .

We need to find a way to put some or all of these together to get:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$ .

So I do notice on the right hand side the $\sin(a+b)$ and the $\sin(a-b)$ .

Let's start there then.

There is a plus sign in between them so let's add those together:

$\sin(a+b)+\sin(a-b)$

$=[\sin(a+b)]+[\sin(a-b)]$

$=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]$

There are two pairs of like terms. I will gather them together so you can see it more clearly:

$=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]$

$=2\sin(a)\cos(b)+0$

$=2\sin(a)\cos(b)$

So this implies:

$\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)$

Divide both sides by 2:

$\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)$

By the symmetric property we can write:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$

3 0

3 years ago