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Licemer1 [7]
3 years ago
10

What is the length of x?

Mathematics
2 answers:
Kamila [148]3 years ago
8 0

Answer:

9.04

Step-by-step explanation:

tan 37 = x/12

0.7536 x 12= x

x = 9.04

user100 [1]3 years ago
3 0
Hope this helps! feel free to clarify

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The domain of the function f(x) = 3x3 is {2, 5}. what is the function’s range?
Crank
F(2) = 24
f(5) = 250

range is {24, 250}
3 0
3 years ago
M(x)=4x+15;m(x)=7<br><br> Answer?
anyanavicka [17]

Answer:

M=0

Step-by-step explanation:

(M)=4M+15(M)=7

We move all terms to the left:

(M)-(4M+15(M))=0

We add all the numbers together, and all the variables

M-(+19M)=0

We get rid of parentheses

M-19M=0

We add all the numbers together, and all the variables

-18M=0

M=0/-18

M=0

6 0
3 years ago
Circle the letter for the correct answer.
8_murik_8 [283]

Answer:

Step-by-step explanation:205

7 0
3 years ago
If h(x) = 4x - 3, what is an equation for h-1(x)?
AleksandrR [38]

Answer:

h(x-1)=4x-7

Step-by-step explanation:

Given :

h(x)=4x-3

Now,

putting the value of x =x-1

h(x-1)=(4(x-1)-3)

h(x-1)=4x-4-3

h(x-1)=4x-7

Answer will be h(x-1)=4x-7

7 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!

<h2>(a)</h2>

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)}

<h2>(b)</h2>

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.

<h2>(c)</h2>

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

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. 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
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