Answer:
320m
Step-by-step explanation
a rectangle has four sides, two pairs of identical sides. So two sides are 16. So that adds up to 32. Perimeter is all sides added together. so it has to add up to 72. 72 minus 32 is 40, so two of the sides have to be 40. Area is length times width, so 16 times 20, which is 320.
Answer:
-15 < 20
Step-by-step explanation:
5(w - 3) < 5w + 20
Distribute;
5w - 15 < 5w + 20
Subtract 5w from both sides;
-15 < 20
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
Step-by-step explanation:
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