<span>-6a^-3 b^9 c
-------------------
18a^2 b^-2 C^4
= a^-1 b^7
----------------
-3 c^3
= b^7
-------------
-3 a^1 c^3</span>
Answer:
<u>Point C</u>, because it is the only place on the picture where the lines cross.
Answer:
The students spent 135 minutes in lecture and 115 minutes in a lab.
Step-by-step explanation:
Let x and y be the respective amount of time students spent in lecture and lab.
x-20=y --- eqn 1
x+y=250 --- eqn 2
From eqn 1, x=y+20 --- eqn 3
Sub eqn 3 into eqn 2:
y+20+y=250
2y=230
y=115
Sub y=115 into eqn 3:
x=115+20
=135
∴x=135, y=115
Answer:
Length is 14
Width is 6
Step-by-step explanation:
Length = 2W + 2
Width = W
Perimeter is twice the length + twice the width
P = 40 = 2(2W+2) + 2W first simplify
40 = 4W + 4 + 2W now combine terms
40 = 6W + 4 and subtract 4 from both sides
36 = 6W
Width W = 6
Length = 2W + 2 = 12 + 2 = 14
The answer is 1
Use the discramnt formula
![{b {}^{2} - 4ac }](https://tex.z-dn.net/?f=%20%20%7Bb%20%7B%7D%5E%7B2%7D%20-%204ac%20%7D%20)
Where a is 81, B is 18, and c is 1
We get
![18 {}^{2} - 4(81)(1)](https://tex.z-dn.net/?f=18%20%7B%7D%5E%7B2%7D%20%20%20-%204%2881%29%281%29)
We then get 324-324=0
Which equal 0. Since the discramnt produce 0, the number of solution is one and repeated.