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2 years ago

Michelle used a $20 bill to pay for a notebook that only costs P dollars

Mathematics

2 answers:

AleksandrR [38]2 years ago

6 0

Answer:

20-p

Step-by-step explanation:

VladimirAG [237]2 years ago

3 0

20 - p <=== this is ur expression for how much change she will get back

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PLZ SHOW ALL WORK

lara31 [8.8K]

Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width: w = 12 - 2x
length: l = 18 - 2x
height: h = x

Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x

Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6

Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160

Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224

And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.

3 0

3 years ago

Why does 30+(-8)+(-6) equal 16? I'm so confused on this one

Daniel [21]

30+(-8)+(-6)
30-8-6
22-6
16

5 0

2 years ago

Read 2 more answers

Solve for n. 9n + 120 = 336

rewona [7]

Answer:

the answer is n=24

Step-by-step explanation:

hope this helps :)

8 0

2 years ago

Read 2 more answers

11. If f(x)=4x-9, what is the equation for f^-1(x)?

FrozenT [24]

F(x) = 4x - 9

let f(x) = y, this implies that x = f⁻¹(y)

y = 4x - 9 Let us solve for x.

4x - 9 = y

4x = y + 9

x = (y + 9)/4

Recall that x = f⁻¹(y),

x = (y + 9)/4

f⁻¹(y) = (y + 9)/4

That means that for f⁻¹(x)

f⁻¹(x) = (x + 9)/4

Hope this explains it.

3 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

3 years ago