On Friday, the temperature was 82°F. The temperature changed by -2°F on Saturday, and then it changed by 2°F on Sunday.
1 answer:
You might be interested in
Answer:
<em>797.0ft²</em>
Step-by-step explanation:
<u>Find a related diagram attached</u>
Area of the gate = Area of the semicircle + Area of the Rectangle
Area of the semicircle = πr²/2
r is the radius
From the diagram, radius r = 20/2 = 10ft
Area of the semicircle = 3.14(10)²/2
Area of the semicircle = 3.14(100)/2
Area of the semicircle = 314/2
Area of the semicircle = 157ft²
Area of the rectangle = Length * Width
Area of the rectangle = 20*32
Area of the rectangle = 640ft²
Area of the gate = 157 + 640
Area of the gate = 797ft²
<em>Hence the area of the gate is 797.0ft²</em>
Problem 5
<h3>Answer: 6i</h3>------------------
Explanation:
We have these four identities
- i^0 = 1
- i^1 = i
- i^2 = -1
- i^3 = -i
Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.
We can think of it like this if you wanted
i^28 = (i^4)^7 = 1^7 = 1
Then the sqrt(-36) becomes 6i
So overall, we end up with the final answer of 6i
=============================================
Problem 6
<h3>Answer: -3i</h3>------------------
Explanation:
We'll use the ideas mentioned in problem 5
46/4 = 11 remainder 2
i^46 = i^2 = -1
sqrt(-9) = 3i
The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i
In other words, here is one way you could write out the steps
=============================================
Problem 7
<h3>Answer: -1</h3>------------------
Work Shown:
i^10 = i^2 because 10/4 = 2 remainder 2
i^19 = i^3 because 19/4 = 4 remainder 3
i^7 = i^3 because 7/4 = 1 remainder 3
Again, all we care about are the remainders.
=============================================
Problem 8
<h3>Answer: -1 + i</h3>------------------
Work Shown:
i^22*i^6 = i^(22+6) = i^28
Earlier in problem 5, we found that i^28 = i^0 = 1
So,