Mathematics 47 Online
OpenStudy (anonymous):

find the value of y. express in simplest radical form.

This question doesn't make sense alone. Is there an equation involved?

OpenStudy (anonymous):

a triangle

OpenStudy (anonymous):

the hypotenuse is $\sqrt{85}$ the base is y and the height is 6

Ok. So if it has a hypotenuse, then it is a right triangle. What equation did we say relates the sides of a right triangle?

OpenStudy (anonymous):

i dont know

The Pythagorean theorem. Check the previous answer I gave you for what that looks like, and then let's continue.

OpenStudy (anonymous):

ok but i never have done it with a radical of 85

Ok, so that's fine. Once we've said that $$a^2 + b^2 = c^2$$, and we know that the hypotenuse is $$\sqrt{85}$$, we have: $y^2 + 6^2 = \sqrt{85}^2$ What's $$\sqrt{85}^2$$?

OpenStudy (anonymous):

Not quite. The square root is the opposite of squaring something. So a square root and a square cancel out.

OpenStudy (anonymous):

so it cancels all the time?

Right. So in $$\sqrt{x}^2 = x$$. In this case, 85. So we have: $y^2 + 6^2 = 85$ Then you can solve for $$y$$.

OpenStudy (anonymous):

y2 + 65 = 85

$$6^2 = 6 \cdot 6$$, which isn't quite 65.

OpenStudy (anonymous):

i mean 36

Right :)

Exactly.

OpenStudy (anonymous):

so now subtract 65 from both sides?

Well, 36 :)

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

y2=49

OpenStudy (anonymous):

$\sqrt{49}$

OpenStudy (anonymous):

7

Pwned. Exactly.

OpenStudy (anonymous):

wow why cant my teacher explain this like this easy way

That's a question I can't answer :)

OpenStudy (anonymous):

haha yea

OpenStudy (anonymous):

shadow what if the radical 19 is not the hypatenuse but its the height or base do i still keep 19 or do i do 19 x 19

When you plug into the quadratic equation, you'll still have it squared. So if it's $$\sqrt{19}$$, then when you square it it will become 19.

OpenStudy (anonymous):

ok thank you do you do trigonometry sin cos and tangent?

Sometimes. I saw your other question, but not sure what you're asking. Add some more details to it if you can.

OpenStudy (anonymous):

where do i find my other question?

OpenStudy (anonymous):

found it