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3 years ago

What percent of 80 is 32? Enter your answer in the box. (Middle School)

Mathematics

1 answer:

Anna007 [38]3 years ago

8 0

Answer:

40 i that is th answer so you should be good and

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Tables

vaieri [72.5K]

The answer isNot proportional

8 0

3 years ago

Read 2 more answers

The number of basketball games that must be scheduled in a league with n teams is given by G(n) = (n2 - n)/2 where each team pla

Pani-rosa [81]

Answer:

There are 7 teams in the league

Step-by-step explanation:

Given that the number of games is given by:

$G(n) = \frac{(n^2-n)}{2}$

The equation will be used to determine the number of games. As we already know, that 21 games are scheduled which means G(n) is 21 so we have to find n when G(n) is 21

Putting this mathematically

$21 = \frac{n^2-n}{2}$

Multiplying equation by 2

$2*21 = 2*\frac{n^2-n}{2}\\42 = n^2-n\\n^2-n-42 = 0\\n^2-7n+6n-42 = 0\\n(n-7)+6(n-7) = 0\\(n-7)(n+6)=0\\n-7 = 0 => n=7\\n+6=0 => n=-6$

As the number of teams cannot be negative, the number of teams in league are 7.

Hence,

There are 7 teams in the league

3 0

3 years ago

Isiah puts a 100-g weight on a pan balance. How many 10-gram weights does he need to balance the scale?

igor_vitrenko [27]

10, because 100 divided by 10 is 10

5 0

3 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

4 0

3 years ago

A pair of equations which has a unique solution x=2 y=-3

nignag [31]

x=2, y=3 is a pair of a equations right there, so a correct solution.

How about another linear system?

Answer: x+y=5, 10x + y = 23

8 0

4 years ago