Question

Find the length of the missing side. Leave your answer in simplest radical form.

Answer 1

Again, Pythagoras.

Answer 2

im finding c?

Answer 3

yes

Answer 4

ah oh so i would set it up like 3^2+4^2
+c^2?

Answer 5

=c^2

Answer 6

exactly

Answer 7

keep in mind that you want c, and not c^2

Answer 8

so then how would i solve it like what would be my first step?

Answer 9

Can you solve such equations:

\[\Large x^3=5 \\ \Large x^2=2 \\ \Large c^9=9 \]

Answer 10

?

Answer 11

noo

Answer 12

I see, well that's a problem (-: I understand now though why you're having trouble.

Answer 13

ok see, x, or c, or whatever variable you want to "isolate" is the base, the number above it is called the exponent.

Answer 14

Whenever you want to 'erase' / get rid/ remove an exponent, all you have to do is taking the root of it. Simple as that, it's not always the square root, for Pythagoras it's always the square root though, but for instance. When you want to solve

\[ \Large x^5=9\]
Then you don't have to take the square root \(\sqrt{....}\) then you have to take the 5th square root, looks like this \( \sqrt[5]{...} \)

Answer 15

See that the Exponents and the Roots you take always match
If you want to erase the exponent 9, you take the 9th square root, if you want to erase the 2, you take the square root. If you want to erase the exponent 3, you take the third square root.

Answer 16

any better/clearer?

Answer 17

yeah i understand that so if it was x^11 i would take the squre root of 11?

Answer 18

well you call it the 11th square root, but I am not sure about the english terminology, but you got it, find that on your calculator, you usually have two buttons. One looks like this:
\[\Huge \sqrt{ \ \ \ } \longrightarrow \text{square root}\]
and the other looks like that:
\[\Huge \sqrt[x]{  \   } \longrightarrow \text{ x'th root}\]

Answer 19

you have to make sure that you feel absolutely comfortable with these operations. If you have troubles with them most problems will remain insolvable, so in case you need help with algebraic operations let me know. Not all of them are simple, and not all of them are necessary for your level (I guess early high school), but the most basic ones, like isolating, you should be able to do that at any given time.

Answer 20

so then for the problem im doing i would take the sq root of it?

Answer 21

exactly, get all numbers on one side, that's easy, now you have something like: \[ \Large c^2=somenumber\]
But you don't want c^2, mathematically c^2 is a square (see square again), so you want to take a square root

Answer 22

so can i put it into the calculator like 4^2+3^2? i got 25 when i did that

Answer 23

exactly and now take the square root of that.

Answer 24

okay i got 5

Answer 25

voila

Answer 26

so what if i got another problem and the number is the c^2 would i solve the same waY?

Answer 27

always, you see mathematics is all about finding structure, what we did yesterday, were the same exact kind of equations then this one.

Answer 28

oh okay im gonna show the pic of the next one and i'll do it and tell me if its right

Answer 29

sounds good to me.

Answer 30

well i got 164 and squared it and got 12.8 but i dont see that as an answer choice did i do something wrong?

Answer 31

Do you know how to name the sides of a triangle ?

Answer 32

yeah leg leg hypotenuse?

Answer 33

exactly, and c^2 is the hypotenuse, can you find the hypotenuse on the triangle above?

Answer 34

Sorry, c is the hypotenuse

Answer 35

ahh i thought thats what i was doing

Answer 36

You don't want to solve for c now, Pythagoras is still valid:

\[ \Large a^2+b^2=c^2 \]

Answer 37

But see, in the picture above you have the hypotenuse already given, the hypotenuse is ALWAYS the longest side of a right-angled triangle, and it is ALWAYS at the other side of the 90degree, so if you see a 90 degree angle, you immediately know that the side opposite of it is the hypotenuse.

Answer 38

oh okay so i got the hypotenuse so im solving for a^2?

Answer 39

not for a^2, but for a, you can also solve for b.. See Pythagoras can also be understood like that:
\[\Large leg_a^2 + leg_b^2=hypotenuse^2 \]

Answer 40

You have given: One leg, and the hypotenuse, therefore you want to solve for the remaining leg, it doesn't matter which one you call how.

Answer 41

okay i sqared 8 and 10 and i got 8^2=2.8 10^2=3.1

Answer 42

did i do it correct?

Answer 43

\[\Large a^2+b^2=c^2 \]
Solve this equation to a for me.

Answer 44

is a varible squared equals 1 or does it just equal the variable ? like c^2=1 or c?

Answer 45

no, a variable squared is just a variable squared. It doesn't necessary equal one.

Answer 46

See
\[\Large a^2=c^2-b^2 \]
All I did was subtracting b from both sides above.

Answer 47

okay so then how could i solve this?

Answer 48

and now, you know how to isolate a. Look at the exponent again .

Answer 49

oh wow i was thinking of something different. okay so i would do -8^2 to both sideS?

Answer 50

then it would be like b^2=10^2-8^2?

Answer 51

yes and now isolate b.

Answer 52

because you want b, not b^2

Answer 53

so not i take the sq root of 8 and 10?

Answer 54

of the expression you evaluated on the RHS (right hand side)

Answer 55

yeah

Answer 56

do that yes.

Answer 57

i got .333 when i put in the calculator?

Answer 58

\[b^2=10^2-8^2=36 \]

Answer 59

so what is b?

Answer 60

6? oh you know what i did i put it in the calculator under sq root instead of squared

Answer 61

6 is the right answer.

Answer 62

I recommend you to practice a bit more on these type of questions. They are easy, once you grasp the concept behind it.

Answer 63

yeah it is i have one more ill show you it and do it

Answer 64

i got 44 and i took the sq root and got 6.6 but i dnt see that as an answer choice?

Answer 65

i did something wrong?

Answer 66

it's correct, maybe they want you keep it as a radical

Answer 67

yeah it says keep in radical form but wouldnt that be 44?

Answer 68

you can always factor it. divide by 4, gives you eleven

Answer 69

so 2sqrt11

Answer 70

ohhh okay

Answer 71

thank you once again lol :) are you like a teacheR?

Answer 72

if you have troubles with the radicals, you can also enter the results they give you to your calculator and just verify for yourself if one of them is equal to yours.
And no, just a full-time nerd, but I tutor Mathematics for a University, first semesters

Answer 73

your not a nerd ! your just really smart, and wow that's great . that university needs you. but thank you again so much for your help :)!

Answer 74

no problem

Answer 75

Can you answer more questions? or are you busy?