Answer: 
![yy = 0.40 (20 - 14) + 14[tex] yy = 2.4 + 14](https://tex.z-dn.net/?f=%20yy%20%3D%200.40%20%2820%20-%2014%29%20%2B%2014%3C%2Fp%3E%3Cp%3E%5Btex%5D%20yy%20%3D%202.4%20%2B%2014%20)

Thus, 16.4 pound class belongs to the 40th percentile of the weight class.
Answer:
d=2√13
Step-by-step explanation:
B(1,4) T(4,6) what is BD
find BT
d=√(x2-x1)²+(y2-y1)²
d=√(4-1)²+(6-4)²
d=√3²+2²
d=√9+4
d=√13 this is haf the distance
BD=2d=2√13
Answer:
Perimeter of PQR = 37 units (Approx.)
Step-by-step explanation:
Using graph;
Coordinate of P = (-2 , -4)
Coordinate of Q = (16 , -4)
Coordinate of R = (7 , -7)
Find:
Perimeter of PQR
Computation:
Distance between two point = √(x1 - x2)² + (y1 - y2)²
Distance between PQ = √(-2 - 16)² + (-4 - 4)²
Distance between PQ = 18 unit
Distance between QR = √(16 - 7)² + (-4 + 7)²
Distance between QR = √81 + 9
Distance between QR = 9.48 unit (Approx.)
Distance between RP = √(7 + 2)² + (-7 + 4)²
Distance between RP = √81 + 9
Distance between RP = 9.48 unit (Approx.)
Perimeter of PQR = PQ + QR + RP
Perimeter of PQR = 18 + 9.48 + 9.48
Perimeter of PQR = 36.96
Perimeter of PQR = 37 units (Approx.)
Answer:
А.The system has two solutions, but only one is viable because the other results in a negative width.
Step-by-step explanation:
Given
Let:
length of play area A
width of play area A
length of play area B
width of play area B
Area of A
Area of B
From the question, we have the following:




The area of A is:

This gives:

Open bracket

The area of B is:


Substitute: 

Open brackets


Expand


We have that:

This gives:

Collect like terms


Using quadratic calculator, we have:
or
--- approximated
But the width can not be negative; So:

B² = 8b + 84
b² - 8b - 84 = 0
b = <u>-(-8) +/- √((8)² - 4(1)(-84))</u>
2(1)<u>
</u>b = <u>8 +/- √(64 + 336)</u>
2
b = <u>8 +/- √(400)
</u> 2<u>
</u>b = <u>8 +/- 20
</u> 2
b = 4 <u>+</u> 10
b = 4 + 10 b = 4 - 10
b = 14 b = -6
<u />