Answer:
Step-by-step explanation:
First do your x2-x1/y2-y1 to find slope, which is 0-3/7-13 = -3/-6 = 1/2
Then plug into point slope form (choose either point)
(y-13) = 1/2(x-0)
OR
(y-7) = 1/2(x-3)
Now, as the pizza is circular, so it does not matter about the box shape. If the box shape increases or decreases, the pizza shape is the same.
Now, we have been asked about the largest pizza radii that can fit in the box, so we will consider the shortest value of the box.
The width is given as 14 inches. There is a 2 inch border around the pizza. So, the width is 14-4 = 10 inches. (Now assume this as the diameter of pizza)
And the radius is =
inches.
So, the largest pizza that can fit in the box will have a radius of 5 inches.
Answer:
Both the parts of this question require the use of the "Intersecting Secant-Tangent Theorem".
Part A
The definition of the Intersecting Secant-Tangent Theorem is:
"If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment."
This, when applied to our case becomes, "The length of the secant RT, times its external segment, ST, equals the square of the tangent segment TU".
Mathematically, it can be written as:
Part B
It is given that RT = 9 in. and ST = 4 in. Thus, it is definitely possible to find the value of the length TU and it can be found using the Intersecting Secant-Tangent Theorem as:
Thus,
Thus the length of TU=6 inches
You can extract two balls of the same colour in two different way: either you pick two black balls or two red balls. Let's write the probabilities of each pick in each case.
Case 1: two black balls
The probability of picking the first black ball is 2/5, because there are two black balls, and 5 balls in total in the urn.
The probability of picking the second black ball is 1/4, because there is one black ball remaining in the urn, and 4 balls in total (we just picked the other black one!)
So, the probability of picking two black balls is

Case 2: two red balls
The probability of picking the first black ball is 3/5, because there are three red balls, and 5 balls in total in the urn.
The probability of picking the second red ball is 2/4=1/2, because there are two red balls remaining in the urn, and 4 balls in total (we just picked the other red one!)
So, the probability of picking two red balls is

Finally, the probability of picking two balls of the same colour is
