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.
Answer:
Kite
Perpendicular
Reflexive
HL
Step-by-step explanation: Just did it on edg 2021.
Part A
The graph is shown below as an attached image.
The diagram shows a straight line that goes through the two points (0,-3) and (1, -5)
I'm using GeoGebra to graph the line.
side note: (0, -3) is the y intercept which is where the graph crosses the y axis.
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Part B
Answer is choice 2
The graph can be written in the form y = mx+b, so it is linear
In this case, m = -2 is the slope and b = -3 is the y intercept
We can write the slope as m = -2 = -2/1. This tells us that we can move down 2 units and then over to the right 1 units to get from point to point. This process of "down 2, over to the right 1" happens when moving from point A to point B in the diagram below.
