Answer:
True
Step-by-step explanation:
Any two sides of a triangle is greater than the third side.
Ans: -4.3
Integers are all whole numbers negative, positive, and 0
Hope this helped :)
so basically what i was thinking of was they sold them in the spring
Answer:
.
Step-by-step explanation:
Since repetition isn't allowed, there would be
choices for the first donut,
choices for the second donut, and
choices for the third donut. If the order in which donuts are placed in the bag matters, there would be
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
times.
For example, if a bag includes donut of type
,
, and
, the count
would include the following
arrangements:
Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count
by
to find the actual number of donut combinations:
.
Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of
objects from a set of
distinct objects:
.
![\huge\boxed{y-5&=-\frac{1}{3}(x+9)}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7By-5%26%3D-%5Cfrac%7B1%7D%7B3%7D%28x%2B9%29%7D)
First, we'll find the slope of the new line. The first line has a slope of
. Take the negative reciprocal of this (Flip the numerator and denominator, then multiply by
) to get
for the new slope.
Then, we'll use the point-slope form to make the new equation, where
is the slope and
is a point on the line:
![\begin{aligned}y-y_1&=m(x-x_1)\\y-5&=-\frac{1}{3}(x-(-9))\\y-5&=-\frac{1}{3}(x+9)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dy-y_1%26%3Dm%28x-x_1%29%5C%5Cy-5%26%3D-%5Cfrac%7B1%7D%7B3%7D%28x-%28-9%29%29%5C%5Cy-5%26%3D-%5Cfrac%7B1%7D%7B3%7D%28x%2B9%29%5Cend%7Baligned%7D)