The water level in a swimming pool increased from 4.5 feet to 6 feet. What is the percent increase in the water rounded to the n
1 answer:
You might be interested in
Answer: 54 cm²
Step-by-step explanation: In this problem, we're asked to find the area of the trapezoid shown. A trapezoid is a quadrilateral with one pair of parallel sides.
The formula for the area of a trapezoid is shown below.
The <em>b's</em> represent the bases which are the parallel sides and <em>h</em> is the height.
So in the trapezoid shown, the bases are 6 cm and 12 cm and the height is 6 cm. Plugging this information into the formula, we have .
Next, the order of operations tell us that we must simplify inside the parentheses first. 6 cm + 12 cm is 18 cm and we have .
is 9 cm and we have 9 cm · 6 cm of 54 cm²
So the area of the trapezoid shown is 54 cm².
<h3>Answer:</h3>
8x = 12
<h3>Explanation:</h3>Subtracting the first equation from the second, you get ...
... (5x +2y) -(3x +2y) = (7) -(5)
... 2x = 2 . . . . . the first answer choice is true
___
Multiplying this expression for x by 4, we get
... 8x = 8 . . . . . the second answer choice (8x=12) is false
___
Adding thee first equation to the second, you get ...
... (3x +2y) +(5x +2y) = (5) +(7)
... 8x +4y = 12
... 8 + 4y = 12 . . . . . . use the value of 8x just computed
... 4y = 4 . . . . . . . . . . subtract 8; the third answer choice is true
_____
<em>Alternate approaches</em>
In short, if you solve the system by any of the methods available, you find x=1 and y=1, so the second answer choice is clearly the correct one:
... 8x ≠ 12
By Cramer's method:
... x = (2·7-2·5)/(2·5-2·3) = 4/4 = 1
... y = (5·5-7·3)/4 = 4/4 = 1
By graphing, see attached.
By substitution (for 2y):
... 5x +(5-3x) =7 . . . . . using 2y=5-3x
... 2x = 2 . . . . . . subtract 5
... x = 1 . . . . . . . .divide by 2
... 5 -3·1 = 2y = 2 . . . . substitute x into the expression for 2y
... y = 1 . . . . . . . divide by 2
By matrix methods, see the second attachment. (x, y) = (1, 1), found in the rightmost column of the result.