You should find that your system doesn't have solutions. This is also confirmed by the fact that the determinant of the matrix of the coefficient of the variables x, y and z is equal to zero.
Answer:
This is a right triangle
Step-by-step explanation:
We will use the Pythagorean theorem
a^2 + b^2 = c^2
If this is true it is a right triangle
5^2 + 12^2 = 13^2
25+144 = 169
169 =169
This is true so this is a right triangle
Answer:
$110
Step-by-step explanation:
Let the fixed amount charged be f and the amount charged to clean a room be r.
To clean six rooms , the company would charge the following, the fixed amount for the call plus the amount charged per room multiplied by 6 which of course is 6r.
Writing this in equation form,
f + 6r = 250
In likewise manner, same applies for 8 rooms.
this gives f + 8r = 320
We now have 2 equations that we will need to solve simultaneously
f + 6r = 250........(i)
f + 8r = 320........(ii)
We subtract equation ii from I to yield
2r = 70, hence r = 70/2 = 35.
We now substitute this in equation 1
f + 6(35) = 250
f + 210 = 250
f = 250 - 210 = 40
Hence, we now know that the amount charged per call is $40 while the amount charged to clean a room is $35.
We now calculate the amount charged by the company to clean 2 rooms.
This is mathematically = f + 2r
40 + 2(35) = $110.
Answer:
Two trapezoids labeled V R A S and B U W D. Sides B U and V R contain one tick mark. Sides B D, U W, V S, and R A each contain two tick marks. Sides A S and W D each contain three tick marks. The angles represented by vertex letters U, B, R, and V each contain two tick marks. The angles represented by vertex letters D, W, S, and A each contain one tick mark.
Answer:
s = height above ground
s = 60 + 20 t - 4.9 t^2 (standard physics equation on earth)
at t = 0
s = 60 (clearly :)
now when does it hit the bleak earth?
That is when s = 0
4.9 t^2 - 20 t -60 = 0
solve quadratic and use the positive t (the negative t was back before you threw it if you had thrown it from the ground)
t = 6.09 or - 2.01
use t = 6.09
now to do the last part there are two obvious ways to get t at the peak
1. look for vertex of parabola
2. look for halfway between t = -2.01 and t = 6.09
I will do it the hard (11) waay by completing the square
4.9 t^2 - 20 t = -(s-60)
t^2 - 4.08 t = -.204 s + 12.2
t^2 - 4.08 t +2.04^2 = -.204 s +12.2 + 4.16
(t-2.04)^2 = -.204(s-80.2)
so
top at 80.2 meters at t = 2.04 s
===============
quick check on time
should be average of 6.09 and -2.01
=4.08 /2 = 2.04 check
