Question

Can someone help me??? please

Answer 1

/wave Afternoon! More fun?

Answer 2

Evaluate the radical notation
\[\frac{ 1 }{ \sqrt[3]{64^2} }\]

Answer 3

@e.mccormick yes! lol, I got some of them done this morning but I got stuck right here.

Answer 4

OK. So you are squaring it and taking the cube root. Good news is it is a power of two and most of those play nice.

Answer 5

so than it will look like:

\[\frac{ 1 }{ \sqrt[3]{4096} }\]

Answer 6

Yes, but you can take the cube root of that. If you are not sure of it, factor nd look for sets of 3 to pull out.

Answer 7

but, what if i just dont want to ? jk

Answer 8

16^3 = 4096 @e.mccormick

Answer 9

Yes, then you do that backwards for the root. So the cube root of 4096 is 16.

Answer 10

@Luis_Rivera But it is math! EVERYBODY wants to do math! Right?

Answer 11

so than would it be \[\sqrt[3]{16}\]

Answer 12

Not quite. You evaaluated the root, so it is gone!

Answer 13

so it would just be \[\sqrt{16} ?\]

Answer 14

@angelina22309 Here is another fact about exponents you may have seen by this point:
\[\sqrt[3]{x}=x^{\frac{1}{3}}\] Now, this lets us solve that last problem a second way that might be easier for you. If you are happy with what you did, great! But if you want a second way, here it is:
\[\frac{1}{\sqrt[3]{64^2}}=\frac{1}{(64^2)^{\frac{1}{3}}}\]
Now, have you seen how exponets to an exponent work? So \[(x^n)^m\]

Answer 15

yea i get that part because the first part to this question was to write 64^-2/3 in radical notation.

Answer 16

No, the root is completley gone, bar and all. \(\sqrt[3]{4096}=16\)
\[\therefore \frac{1}{\sqrt[3]{4096}}=\frac{1}{16}\]

Answer 17

oh I see!! So the answer is 1/16

Answer 18

OK, that \((x^n)^m\) thing is pretty easy to show. Want to learn it real fast?

Answer 19

yes please

Answer 20

Good! Lets use 2 since we have it here in the problem... well, it is hiding in there.
\[4096=2^{12}\] So I am going to show you something else that gets the same number.
\(2^3\) means \(2\cdot 2\cdot 2\), right? Well, \((2^3)^4=(2^3)\cdot (2^3)\cdot (2^3)\cdot (2^3)\)
But since we know what \(2^3\) means, we can put that in and get:
\((2^3)^4=(2\cdot 2\cdot 2)\cdot (2\cdot 2\cdot 2)\cdot (2\cdot 2\cdot 2)\cdot (2\cdot 2\cdot 2)\)
By rules of multiplicaiton we can remove the ( ), so:
\((2^3)^4=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\)
Now that is 12, 2s multiplied, \(\therefore (2^3)^4=2^{12}\)

Answer 21

I get that part.

Answer 22

This is an application of a rule of exponents that says these three things are the same:
\[(x^n)^m=x^{n\cdot m}=(x^m)^n\]
AKA: an exponent to an exponent is the product of the exponents, and I can do products in different orders and get the same results.
So now, if I did this in the problem, I could change it.

Answer 23

ok that makes more sense!

Answer 24

\[\frac{1}{(64^2)^{\frac{1}{3}}}=\frac{1}{(64^{\frac{1}{3}})^{2}}\]
Now I see that the cube root of 64 is 4. That changes this to:
\[\frac{1}{(64^{\frac{1}{3}})^{2}}=\frac{1}{(4)^{2}}=\frac{1}{16}\]

Answer 25

So that is a second way to do roots of powers. If you are careful with the order of oprations, this can let you pick what you do when. With 64 ad 4096, this is useful because 64 is a lot easier to know the cube root of than 4096 is!

Answer 26

Yea! It is lol.
Hey can you check this one for me? I think I did it right...

Find the maximum y-value on the graph of y = f(x)

\[f(x) = -x^2 + 8x + 5\]
\[\frac{ -b }{ 2a } = \frac{ -8 }{ 2(-1) } = 4\]
\[f(4) = -4^2 + 8(4) + 5 = 16 + 32 + 5 = 53\]
Maximum y-value is 53?

Answer 27

The 4 is right.... but lets tak another look at the y of 4.

Answer 28

You made one mistake.
\(-x^2\ne (-x)^2\)
\(-x^2= -(x^2)\)

Answer 29

So you dropped a sign. I think I have dropped a few million by now, so don't be surprised if you drop more yourself!

Answer 30

Wait where did I drop the sign?? @e.mccormick

Answer 31

You:
\(f(4) = -4^2 + 8(4) + 5 = 16 + 32 + 5 = 53\)
Me:
\(f(4) = -(4^2) + 8(4) + 5 = -16 + 32 + 5 = ??\)

Answer 32

But x = -1 so since we are putting 4 as x, wouldnt it be -4^2

Answer 33

That means your answer would be 21

Answer 34

At that point it is 4, and yes, 21. And then to confiirm:
https://www.desmos.com/calculator/a5kixezwea

Answer 35

so my answer is wrong?

Answer 36

You were correct in getting 4 and evaluating \(f(4)\). The mistake was in the meaning of \(-x^2\).
PEMDAS
Powers/Exponents
Multiplication/Division
Addition/Subtraction

\(-x^2\) means \((-1)x^2\). That means you do the exponent first, then the multiplication by \(-1\).

Answer 37

oh ok!! Makes sense

Answer 38

Yah, it is maxed at 21, not 53.

Answer 39

Yah, at that moment you were "thinking x is negative four and square it" by doing the - first. Very, evry easy mistake to make. But because it is a variable, you have to be careful of the order of operations.

Answer 40

Can you help me find the vertex of a problem??

Answer 41

Sure. Need to look at a ref on that, refresh my parabolas, but should not be too hard.

Answer 42

Ah, \(y = a(x – h)^2 + k\) form. OK. I can work things to that easy enough.

Answer 43

For \[y = x^2 - 6 \]

do the following:
a.) Sketch a graph
b.)identify the vertex
c.) Compare the graph of y = f(x) to the graph y = x^2

Answer 44

I've already sketched the graph, however, not sure what the vertex is.

Answer 45

OH! As in what the vertex means?

Answer 46

I know the vertex are coordinates in the graph right?

Answer 47

A parabola can be broken down into parts. There is the curve, the focus, the vertex, and the directrix. Most people know what the curve is. A U shape! At the very, very lowest point of the u, that is the Vertex. Now, if the U were upside down, it would be the highest point.
To be complete, the Focus and Directrix are NOT on the parabola. They are mathematical constructs that tell us about the parabola. If you have not heardof them yet, do not worry about them.

Answer 48

hold on, let me attach my graph

Answer 49

An animated parabola with the vertex as a dot:
http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html

Answer 50

Remember that high point we found? That was one way of finding the vertex.

Answer 51

So the low point on your graph is the Vertex.

Answer 52

vertex = (0, -6)

Answer 53

Yep! The term bertex gets reused in math. If you have an angle, it is the point they meet at. In a parabola, it is where the curve changes from down to up, up to down, etc.

Answer 54

Later you will do a section of math called the conics, as in slices of a cone. The parabola, circle, ellipse, and hyperbola will all be gone over in a lot more deatail. So learning the parabola really well here will help you later.

Answer 55

ok so not onto quadratic equations! Lol I always get stuck in the very end. I'll show you what ive done so far.

\[y = x^2 - 2x - 4\]
\[a = 1, b = 2, c = -4\]

\[x = \frac{ -2 \pm \sqrt{2^2 - 4(1)(-4)} }{ 2(1) }\]
\[x = \frac{ -2 \pm \sqrt{4 + 16} }{ 2 } = \frac{ -2 \pm \sqrt{20} }{ 2 }\]

Answer 56

\(a=1,b=2,c=−4\) Oops! Sign! \(a=1,b=-2,c=−4\)

Answer 57

ahhh!!!! hold on

Answer 58

Beacuse the \(b^2\) that only changes the outside b to positive. Nothing else.

Answer 59

okay so the outside 2 would just be 2 not -2

Answer 60

Yep. Then you get to work th root a bit.\[x = \frac{ 2 \pm \sqrt{4 + 16} }{ 2 } = \frac{ 2 \pm \sqrt{20} }{ 2 }=\frac{ 2 \pm \sqrt{2\cdot 2\cdot 5} }{ 2 }\]

Answer 61

\[x = \frac{ 2 \pm \sqrt[2]{5} }{ 2 }\]

Answer 62

Ummm.... where did the 2 go? It eventually leaves, but it left without its friends. They all stay or the all go (cancel.)

Answer 63

wouuld it look like \[x = \frac{ 2 \pm 2\sqrt{5} }{ 2 }\]

Answer 64

Yes, but now you have a 2 that can factor.

Answer 65

\[x = \frac{ 2 \pm 2\sqrt{5} }{ 2 }\Rightarrow x = \frac{ (2 )(1 \pm \sqrt{5}) }{ (2)(1) }\]

Answer 66

so will it be \[x = 1 + \sqrt{5}, x = 1 - \sqrt{5}\]

Answer 67

Yes!

Answer 68

That last part always gets me!

Answer 69

Yah, the root work. It gave you trouble earlier with the first one we did today.

Answer 70

So we know where the root of the problem is.... roots.

Answer 71

yes!!

Answer 72

I remember doing a problem like this one that ima show you before but I had a lot of trouble with it for some reason . ima close this question and open another .

Answer 73

So I suggest you take sme radical actions!
http://www.purplemath.com/modules/radicals.htm

Review that, make some notes to yourself, and dig out those roots from where they are hiding and into the light understanding!

Answer 74

thanks!

Answer 75

And there are serveral jokes there, especially if you know philosophy. Socrates saw true knowlege as the light of the Sun. A mathematical root is a radical.