Answer:
2000 L
Step-by-step explanation:
There are 1250 L of water in a tank at present. If the tank is 0.625 full, what is the capacity of the tank?
The simple solution is:
1250 L ÷ 0.625 = 2000 L
The algebraic solution is:
Let <em>c</em> equal the capacity of the tank.
Therefore, <em>c</em> × 0.625 = 1250.
Divide both sides by 0.625:
<em>c</em> × 0.625 ÷ 0.625 = 1250 ÷ 0.625
And simplify:
<em>c</em> = 1250 ÷ 0.625
<em>c</em> = 2000
The answer is going to be6/8
Answer:
A
Step-by-step explanation:
Noting that i² = - 1
Given
(7 - 3i)(8 + 4i) ← expand factors using FOIL
= 56 + 28i - 24i - 12i²
= 56 + 4i - 12(- 1)
= 56 + 4i + 12
= 68 + 4i → A
Answer:
Your measurements; Area = 216.108 cm²
Another student's measurements; Area = 216.9404 cm²
- Difference in area could be as a result of human error or perhaps that they made use of different measuring tools.
Step-by-step explanation:
For Your measurements;
Length of rectangle = 20.70 cm
Width of rectangle = 10.44 cm
Area of rectangle is given by; A = length × width = 20.7 × 10.44 = 216.108 cm²
For Another student's measurements;
Length of rectangle = 20.74 cm
Width of rectangle = 10.46 cm
Area = 20.74 × 10.46
Area = 216.9404 cm²
The areas they both obtained are not of equal values and this could be as a result of human error or perhaps that they used different measuring tools.
Consider the equation y = x^2. No matter what x happens to be, the result y will never be negative even if x is negative. Example: x = -3 leads to y = x^2 = (-3)^2 = 9 which is positive.
Since y is never negative, this means the inverse x = sqrt(y) has the right hand side never be negative. The entire curve of sqrt(x) is above the x axis except for the x intercept of course. Put another way, we cannot plug in a negative input into the square root function for this reason. This similar idea applies to any even index such as fourth roots or sixth roots.
Meanwhile, odd roots such as a cube root has its range extend from negative infinity to positive infinity. Why? Because y = x^3 can have a negative output. Going back to x = -3 we get y = x^3 = (-3)^3 = -27. So we can plug a negative value into the cube root to get some negative output. We can get any output we want, negative or positive. So the range of any radical with an odd index is effectively the set of all real numbers. Visually this produces graphs that have parts on both sides of the x axis.
