Answer:
Number of two point goals = 169
Number of three point goals = 123
Number of free throws = 139
Step-by-step explanation:
Let the number of two point goals = ![x](https://tex.z-dn.net/?f=x)
Let the number of three point goals = ![y](https://tex.z-dn.net/?f=y)
Let the number of free throws = ![z](https://tex.z-dn.net/?f=z)
As per the given statements:
![x = 2y-77 ...... (1)](https://tex.z-dn.net/?f=x%20%3D%202y-77%20......%20%281%29)
![z = x-30 ...... (2)](https://tex.z-dn.net/?f=z%20%3D%20x-30%20......%20%282%29)
Using equation (1):
...... (3)
Total number of points from two point goals = ![2x](https://tex.z-dn.net/?f=2x)
Total number of points from three point goals = ![3y](https://tex.z-dn.net/?f=3y)
Total number of points from free throws = ![z](https://tex.z-dn.net/?f=z)
![2x+3y+z=846](https://tex.z-dn.net/?f=2x%2B3y%2Bz%3D846)
using (1) and (3):
![2(2y-77)+3y+2y-107=846\\\Rightarrow 4y-154+5y-107=846\\\Rightarrow 9y-261=846\\\Rightarrow 9y=846+261\\\Rightarrow 9y=1107\\\Rightarrow y =123](https://tex.z-dn.net/?f=2%282y-77%29%2B3y%2B2y-107%3D846%5C%5C%5CRightarrow%204y-154%2B5y-107%3D846%5C%5C%5CRightarrow%209y-261%3D846%5C%5C%5CRightarrow%209y%3D846%2B261%5C%5C%5CRightarrow%209y%3D1107%5C%5C%5CRightarrow%20y%20%3D123)
By (1):
![x = 2\times 123-77 =169](https://tex.z-dn.net/?f=x%20%3D%202%5Ctimes%20123-77%20%3D169)
By (3):
![z =2\times 123-107 = 139](https://tex.z-dn.net/?f=z%20%3D2%5Ctimes%20123-107%20%3D%20139)
So, answers are:
Number of two point goals = 169
Number of three point goals = 123
Number of free throws = 139
Answer:
A. f(x) = −2|x| + 1
Step-by-step explanation:
Answer:
For a circle of radius R, the circumference is:
C = 2*pi*R
where pi = 3.14
And if we have an arc defined by an angle θ, the length of the arc is:
A = (θ/360°)*2*pi*R
Here we can not see the image, then i assume that B is the angle that defines the arc AC.
Now we know that the circumference is 120 in, then:
2*pi*R = 120in
Then the length of the arc is:
A = (θ/360°)*120 in
Then if the angle is 18°, we have:
A = (18°/360)*120 in = 6in
A.
ratio 1 to 1
alright
find the distance between the x values and the y values and seperate each into that ratio
1:1
A to B is (6,12) to (15,-4)
disatnce from 6 to 15 is 9, ratio would be 4.5:4.5=1:1
distance from 12 to -4 is 16, ratio would be 8:8=1:1
so the point would be (4.5,8)
b.
5:2
5+2=7
alright
A to C
(6,12) to (20,12)
distance from 6 to 20 is 14, 14/7=2, 2 times 5=10
distance from 12 to 12=0, so same coordinate
the point is (10,12)
c.
2+3=5
C to B
C is (20,12) and B is (15,-4)
distance from 20 to 15 is 5, so 2 is the x value
distance from 12 to -4 is 16, 16/5 times 2=32/5
the point is (2,32/5)
2/3 = 66%
2/4 = 50%
Therefore, 2/3 is greater