How do I write this in Point Slope, Slope-Intercept, and in Standard form. Thanks for the help!

Mathematics

1 answer:

AnnZ [28]3 years ago

6 0

Point slope form: y - y1 = m (x - x1)
-------------------------------------------------

first you need to find the slope(m) by using the slope formula (seen below)

y2 - y1
m= ---------
x2 - x1

9 - 15
m= ---------
1 - (-2)

-6
m= --------- = -2
3

then choose either one of the coordinates you'd like and plug it in.

Im gonna use (-2, 15)

(y - 15) = -2 (x - (-2))
or
(y - 15) = -2 (x + 2)

slope intercept form: y = mx + b
-----------------------------------------------

you can use the slope formula to find that the slope(m) is -2

y = -2x + b

you can then plug in a coordinate to find b (the y-intercept)

I'll use (-2,15)

15 = -2(-2) +b
15 = 4 + b
-4 -4
---------
11 = b

in conclusion:

the slope(m) = -2

when (-2,15) is used
x = -2
y = 15
b = 11

standard form: ax + by = c
----------------------------------------

using the same coordinate (-2,15)

-2a + 15b = c

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A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

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Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .