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Step2247 [10]
3 years ago
14

What is the slope of the line that runs through points (-7,-20) and (12,18)?  Write the equation of that line.

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

The slope is 2 which means the equation would be y=2x.

Step-by-step explanation:

gavmur [86]3 years ago
3 0

Answer:

2

Step-by-step explanation:

-20 - 18 = -38

-7 - 12 = -19

-38/-19 = 2

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A theater can seat 170 people. The number of rows is 7 less than the number of seats in each row. How many rows of seats are the
maks197457 [2]
Answer:number of rows =10
number of places in row = 17

Step-by-step explanation:
number of rows =N
number of places in row = N+7
N(N+7)=170
N^2+7n-170=0
(N+17)(N-10)=0
N1=-17
N2=10
We can`t take negative value for quantity of smth so we throw N1 away
From that number of rows =10
number of places in row = 10+7=17
5 0
2 years ago
I need help pls answer right and don't just pick an answer again pls help.
lara31 [8.8K]

Answer:

-4

Step-by-step explanation:

It the slope hits the y axis, that number will be the y intercept

3 0
3 years ago
Let F: R → R be defined by f(x) = x”. Show that f'is one-to-one and onto.​
Brrunno [24]

Answer:

both

Step-by-step explanation:

3 0
2 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
Which of the following is equivalent to (2x+3)(x-7)?
maw [93]
I’m pretty sure it’s it’s between f and g. Sorry if I’m not exact but hope this helps
8 0
3 years ago
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