For this case we find the slopes of each of the lines:
The g line passes through the following points:
So, the slope is:
Line h passes through the following points:
So, the slope is:
By definition, if two lines are parallel then their slopes are equal. If the lines are perpendicular then the product of their slopes is -1.
It is observed that lines g and h are not parallel. We verify if they are perpendicular:
Thus, the lines are perpendicular.
Answer:
The lines are perpendicular.
Answer:
length and width=4
height=8
Step-by-step explanation:
Hello to solve this problem we must propose a system of equations of 3x3, that is to say 3 variables and 3 equations.
Ecuation 1
Leght=Width
.L=W
Ecuation 2
To raise the second equation we consider that the length and width of 4 inches less than the height of the box
H-4=W
Ecuation 3
To establish equation number 3, we find the volume of a prism that is the result of multiplying length, width, and height
LxWxH=128
From ecuation 1(w=h)
solving for H
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<em>Using ecuation 2</em>
H-4=W
Now we find the roots of the equation, 2 of them are imaginary, and only one results in 4
W=4in
L=4in
to find the height we use the ecuation 2
H-4=W
H=4+W
H=4+4=8
H=8IN
Answer:
6 clown fish and 4 angel fish
Step-by-step explanation:
<u>lets say he bought 8 clown fish. </u>
12*8 = 96
the last 4 dollars wont be used.
<u>lets say he bought 7 clown fish-</u>
12*7 = 84
100 - 84 = 16. 16 is not a multiple of 7.
<u>lets say he bought 6 clown fish,</u>
12*6 = 72
100 - 72 = 28
28/7 = 4
If David has $100 dollars to spend and wants to spend every dollar, he can buy 6 clown fish and 4 angel fish.
If feet:
34' 4" - 8' = 26' 4"
if inches: = 33' 8"
The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
