Answer:
3/28
Step-by-step explanation:
You can rewrite 3/8 ÷7/2 to 3/8 x 2/7
then you can simplify which is 3/28
Answer:
As a fraction:
x | y
________
|
7/3 | 2
3 | 4
11/3 | 6
13/3 | 8
As a mixed fraction:
x | y
________
|
2 ⅓ | 2
3 | 4
3 ⅔ | 6
4 ⅓ | 8
As a decimal:
x | y
___________
|
2.33.. | 2
3 | 4
3.66.. | 6
4.33.. | 8
You should start with x values since the y values will be easier to plot. x's will not be easy to represent.
For example:
x | y
________
|
0 | -5
1 | -2
2 | 1
3 | 4
4 | 7
5 | 10
Answer:
468.75 square inches
Step-by-step explanation:
Let the length of Julian's tank be x
Let the width of Julian's tank be y
So, Area of Julian's tank=
Since we are given that Julian's tank is approximately 300 square inches
So, 
--1
Now we are given that The dimensions of his friend's tank are each exactly
times the dimensions of Julian's tank.
So,length of his friend's tank =
So,width of his friend's tank =
So, Area of his friend's tank=
=
=
(Using 1)
=
Hence the approximate area of the rectangular base of his friend's tank is 468.75 square inches.
So, Option D is correct.
Problem 18)
The distance from J to L is 30 units. We can count out the number of spaces or do subtraction to get 30-0 = 30. I subtracted the number line coordinate of J and L. The result is positive as distance is never negative.
Take 2/3 of this value to get (2/3)*30 = 20 which means that we start at J and add on 20 units to get 0+20 = 20
So the final answer is 20, which is where this point is located on the number line.
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Problem 19)
The distance from A to B is 3 units because |-5-(-2)| = |-5+2| = |-3| = 3
Double this value to get 2*3 = 6
Now add 6 units to the coordinate of point A to get
-5+6 = 1
The location of this point is at 1 on the number line.
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Problem 20)
From J to I, or vice versa, we travel 5 units.
Multiply this by 1.5 to get 1.5*5 = 7.5
So we subtract 7.5 units from 0 (the coordinate of point J) getting us 0-7.5 = -7.5
Final Answer: -7.5
Note: this is the midpoint of -10 and -5 on the number line
