Answer:
Addition but... 40/100x1540 = 616.00
Step-by-step explanation:
edit: i goofed
Answer:
Width of rectangle = 4 ft
Step-by-step explanation:
Given:
Area of rectangle = 48 ft²
Length of rectangle = 3[Width of rectangle]
Find:
Width of rectangle
Computation:
Assume;
Width of rectangle = a
So,
Length of rectangle = 3[Width of rectangle]
Length of rectangle = 3a
Area of rectangle = l x b
48 = 3a x a
3a² = 48
a² = 16
a = 4
Width of rectangle = 4 ft
Length of rectangle = 3[Width of rectangle]
Length of rectangle = 3[4]
Length of rectangle = 12 ft
The answer is
![2 \frac{1}{6}](https://tex.z-dn.net/?f=2%20%20%5Cfrac%7B1%7D%7B6%7D%20)
.
Explanation:
In order to subtract the fractions, we must make them like fractions. To do this, the denominators must be the same by multiplying (only). Since the first fraction is
![\frac{10}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B10%7D%7B3%7D%20)
, 3 can be multiplied by 2 to get 6, which is the other fraction, we can multiply it. Whatever you do to the denominator you must do to the numerator. Now multiply 2 by the numerator (10) to get
![\frac{20}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B20%7D%7B6%7D%20)
. Now we can subtract the fractions
![\frac{20}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B20%7D%7B6%7D%20)
and
![\frac{7}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B7%7D%7B6%7D%20)
to get 13/6. Since this fraction is not in mixed fraction form yet, we must do that first. goes into 13 twice, so the whole number is 2 and there is still 1 left, making the fraction
![\frac{1}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B6%7D%20)
. Therefore, the difference is 2
![\frac{1}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B6%7D%20)
.
Factorize the sum of cubes:
![x^3 + y^3 = (x+y) (x^2 - xy + y^2)](https://tex.z-dn.net/?f=x%5E3%20%2B%20y%5E3%20%3D%20%28x%2By%29%20%28x%5E2%20-%20xy%20%2B%20y%5E2%29)
so
is a factor of
. Then the HCF is
.