Answer: 0
.
0
0
0
1
2
−
0
.
0
2
+
5
+
5
0
0
Step-by-step explanation:
The answer is 4 3/8 boxes because 1 3/4 x 2 1/2 = 4 3/8.
I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.
Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:
a = 1
b = 2
c = 3
d = 4


Simplify into standard form:



Use the quadratic formula to solve:

For functions in the form of

. So in this case:
a = 1
b = -4
c = 2
Plug them in:

Solve for 'x':




So the answer would be A.
<h2>
The area of a triangle is =54 square units</h2><h2>
The perpendicular distance from B to AC is = 
</h2>
Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

The area of a triangle is= ![\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bx_1%28y_2-y_3%29%20%2Bx_2%20%28y_3-%20y_1%29%2Bx_3%28y_1-y_2%29%5D)
=![|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|](https://tex.z-dn.net/?f=%7C%5Cfrac%7B1%7D%7B2%7D%20%5B2%282-8%2B12%288-1%29%2B12%281-2%29%5D%7C)
=
= 54 square units
The length of AC = 
= 
=
units
Let the perpendicular distance from B to AC be = x
According To Problem

⇔
units
Therefore the perpendicular distance from B to AC is = 
Answer:
2.50
Step-by-step explanation:
30% of 25 is 7.50 take that away from 25 gives you 17.50 take that away from 20 gives you 2.50