To find any side of a right angled triangle, we use Pythagorean Theorem. It states that the square of hypotenuse is equal to the sum of squares of it's sides.

Let hypotenuse is a

Perpendicular is b

and base is c

According to the theorem:

a²= b² + c²

Add values and solve for the unknown variable / side.

If you had added the problem, I would have elaborated more but still this will help you out.

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3 years ago

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Use vertical multiplication to find the product of:

Rom4ik [11]

Answer:

$x^6+x^4+4x^3-2x^2-x+3$

Step-by-step explanation:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

First step multiply your terms in your first expression just to the 1 in the second expression like so:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$ Anything times 1 is that anything.

That is, $(x^3+2x+3) \cdot 1=x^3+2x+3$ .

Now we are going to take the top expression and multiply it to the -x in the second expression. $-x(x^3+2x+3)=-x^4-2x^2-3x$ . We are going to put this product right under our previous product.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$

$-x^4-2x^2-3x$

We still have one more multiplication but before we do that I'm going to put some 0 place holders in and get my like terms lined up for the later addition:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

Now for the last multiplication, we are going to take the top expression and multiply it to x^3 giving us $x^3(x^3+2x+3)=x^6+2x^4+3x^3$ . (I'm going to put this product underneath our other 2 products):