What is the midpoint of the line segment wit endpoints (5,6) and (-2,-6)?

What is the answer? I find this very confusing for me.

vichka [17]

Answer:

19

Step-by-step explanation:

If you want to find the answer of the equation, sub the given information into it:

z + xz + y

we know that:

x = 5

y = 1

z = 3

We got:

= 3 + 5*3 + 1

= 3 + 15 + 1

= 19

Hope this helped :3

5 0

3 years ago

Anybody know the answer I need help. Please

Evgen [1.6K]

Answer:

$\dfrac{p^2+9p+2}{p^2-49}$

Step-by-step explanation:

Problems like this require that you recognize that the denominator of the right term is a factor of the denominator of the left term. That is, you're supposed to know how to recognize and factor the difference of two squares.

$\dfrac{5-p}{49-p^2}+\dfrac{p+1}{p-7}=\dfrac{p-5}{p^2-49}+\dfrac{p+1}{p-7}\\\\=\dfrac{p-5}{(p-7)(p+7)}+\dfrac{p+1}{p-7}\cdot\dfrac{p+7}{p+7}\\\\=\dfrac{(p-5)+(p+1)(p+7)}{(p-7)(p+7)}=\dfrac{p-5+p^2+8p+7}{p^2-49}=\dfrac{p^2+9p+2}{p^2-49}$

8 0

3 years ago

A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang

laiz [17]

Answer-

a. Probability that three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

Solution-

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

$\binom{19}{3} = \frac{19!}{3!16!} = 969$

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

$\binom{33}{6} = \frac{33!}{6!27!} =1107568$