Answer:
The width of rectangle w = 4 units
Step-by-step explanation:
We are given:
Width of rectangle = w
Length of rectangle = w+ 4
Area of rectangle = 32 units
We need to find width of the rectangle
The formula used is:
Putting values and finding width of rectangle
We need to solve this quadratic equation to find width (w)
We will use factorisation and break the middle term.
Since width of rectangle can't be negative, so rejecting w=-8
Therefore, the width of rectangle w = 4 units
Answer:
or 6/5
Step-by-step explanation:
72/60
= 6/5
= 1 1/5
= 1.2 (in decimal)
Answer:
Option A.
Step-by-step explanation:
Note: Let as consider, we have to find the total amount after 9 years.
It is given that,
Principal amount = $1000
Rate of compound (yearly) interest = 15% = 0.15
Time = 9 year
The formula for total amount is
where, P is principal, i is rate of interest and n is number of years.
Substituting P=1000, i=0.15 and n=9, we get
So, the total amount after 9 years is $3517.88.
Therefore, the correct option is A.
In general, a rhombus' diagonals have different lengths. This limits the rotational symmetry order to be 2, namely, it gets back to the same shape by rotating 180 degrees. There is a special kind of rhombus where the diagonals have equal lengths, in which case the rhombus has order of rotational symmetry of four. We also call this special kind of rhombus a square.
Answer:
12 bats
Step-by-step explanation:
Let us number the bat caves from 1 to 45 and divide them into 5 parts:
a) 1
b) 2 to 29 ( bats)
c) 30
d) 31 to 44 ( bats)
e) 45
It is given that any 7 caves in a row has 77 bats. Hence, the total number of bats in part b and d = ( ) + ( ) = 462 bats.
There remains ( 490 - 462 ) = 28 bats in cave number 1, 30 and 45.
Now, our question demands the maximum possible number of bats in cave 30.
Minimum number of bats in a cave is 2. So we shall put 2 bats in the last cave, which gives us bats in the first cave.
Therefore, the greatest possible number of bats in cave number 30 = ( 28 - 2 - 14 ) = 12 bats.