What the above statement is saying is:
If the number is 4, find the difference between the square of the number (the number is 4) and the number itself (the number is still 4).
The square of the number is:
The number^2
= 4^2
= 4 × 4
= 16
The difference between the square of the number and the number is:
The number^2 - The number
= 16 - 4
= 12
Hope this helps! :)
Answer:
I just had this problem on a test! I hope this helps, i will attach a picture of my messy work, but hopefully that will give you an idea on how do do it!
How to Solve!-
Repeat this distance formula for each side
Answer:
$ 2330
Step-by-step explanation:
a = $240
Earning at the end of fifth week = $ 680
increase in earnings per weak = ( 680 -240)/2 = 440/2 =110
d =110
l = a + (n-1)d
a20 = 240 + (19 * 110 ) = 240 + 2090 = 2330
Answer:
Volume = 12 *6 *8 , Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 )
Step-by-step explanation:
Given : Cuboid with length 12 , width 6 and height 8 units.
To find : Drag each expression to show whether it can be used to find the volume, surface area, or neither.
Solution : We have given Cuboid with
Length = 12 units ,
Width = 6 units
Height = 8 units.
Volume of cuboid = length * width * height .
Volume = 12 *6 *8.
Surface area = 2 ( l *w + h *+w *h)
Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 ).
None = 12 +6 +8.
Therefore, Volume = 12 *6 *8 , Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 ) .
Answer:
40 square metres
Step-by-step explanation:
The shaded region is of a triangle, whose area is denoted by: A = (1/2) * b * h, where b is the base and h is the height.
Since the left figure is a square with side lengths 10, we know that the height of the triangle is also 10 metres. The right figure is a rectangle with length 4. Since the total base length of the entire figure is 18 and the base of the square is 10, then the width of the rectangle is 18 - 10 = 8 metres.
This width is also the base of the triangle, so b = 8.
Now plug these values into the equation:
A = (1/2) * b * h
A = (1/2) * 8 * 10 = (1/2) * 80 = 40
The area is 40 square metres.