Which statements are correct about quadrilaterals?
1 answer:
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Answer:
Perimeter: 52 Area: 181-16i?
Step-by-step explanation:
<u>Perimeter</u>
The formula for perimeter is 2l+2w, so you plug in the binomials they give you:
2(11-4i)+2(15+4i)
expand:
22-8i+30+8i
combine like terms (which cancels out 8i and -8i):
22+30=52
<u>Area</u>
"i" is not a real number and thus should not be part of the area here, meaning this answer as it is is wrong and I don't know how one would change that.
Here's what I did to get 181-16i, in any case:
The formula for area is l*w, so I plugged in the numbers:
(11-4i)(15+4i)
expanded:
165+44i-60i-16i*i
and combined like terms:
181-16i
*i is , so
=-1 and -1*-16=16.
Answer:
-3 1/3
Step-by-step explanation:
The quadratic
... y = ax² +bx +c
has its extreme value at
... x = -b/(2a)
Since a = 3 is positive, we know the parabola opens upward and the extreme value is a minimum. (We also know that from the problem statement asking us to find the minimum value.) The value of x at the minimum is -(-4)/(2·3) = 2/3.
To find the minimum value, we need to evaluate the function for x=2/3.
The most straightforward way to do this is to substitue 2/3 for x.
... y = 3(2/3)² -4(2/3) -2 = 3(4/9) -8/3 -2
... y = (4 -8 -6)/3 = -10/3
... y = -3 1/3
_____
<em>Confirmation</em>
You can also use a graphing calculator to show you the minimum.
