Y=Acos(p)+m, A=amplitude, p=period, m=midline, in this case:
A=1/2, p=360(t/12)=30t, m=(10-9)/2+9=9.5 so
h(t)=(1/2)cos(30t)+9.5
The value of 'x' is 24.2 and the value of 'y' is 46.5.
To solve this, we do the following steps.
<u>Step 1:</u> Divide 'y' into 2 parts, 'a' and 'b'. 'a' would be the lower leg of the 45°-45°-90° triangle, while 'b' is the lower leg of the 30°-60°-90° triangle.<em>
</em><u>Step 2:</u> Given the hypotenuse (34) of the 30°-60°-90° triangle, solve for 'b' using the cosine of 30°.
cos30° = b/34 [adjacent over hypotenuse]
b = 34cos30° [cross-multiply]
b = 29.4
<u>Step 3:</u> Solve for the 90° leg (the side opposite the 30° angle) using the Pythagorean Theorem. We will name this leg "h" (cuz height).
l² + l² = hyp²
29.4² + h² = 34²
h² = 1156 - 864.36
√h² = √291.64
h = 17.1
<u>Step 4:</u> Solve for 'x' by using the 45°-45°-90° triangle ratio (1:1:√2). √2 would be the hypotenuse of the 45°-45°-90° triangle, while 1 would be both congruent legs.
Side 'h' is one of the legs; side 'a' is the other. Since these legs are congruent, 'a' also measures 17.1. Now all we need to do is solve for 'x', which is our hypotenuse. To do this, we simply multiply the measure of side 'h' or 'a' by √2.
x = 17.1 × √2
x = 24.2
<u>Step 5:</u> Now that we got the value of 'x', solve for 'y' by adding the measures of sides 'a' and 'b' together.<em>
</em><u /> y = a + b
y = 17.1 + 29.4
y = 46.5
And there you have it! <em>Hope this helps.</em>
<em>
</em>
Problem 18)
The distance from J to L is 30 units. We can count out the number of spaces or do subtraction to get 30-0 = 30. I subtracted the number line coordinate of J and L. The result is positive as distance is never negative.
Take 2/3 of this value to get (2/3)*30 = 20 which means that we start at J and add on 20 units to get 0+20 = 20
So the final answer is 20, which is where this point is located on the number line.
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Problem 19)
The distance from A to B is 3 units because |-5-(-2)| = |-5+2| = |-3| = 3
Double this value to get 2*3 = 6
Now add 6 units to the coordinate of point A to get
-5+6 = 1
The location of this point is at 1 on the number line.
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Problem 20)
From J to I, or vice versa, we travel 5 units.
Multiply this by 1.5 to get 1.5*5 = 7.5
So we subtract 7.5 units from 0 (the coordinate of point J) getting us 0-7.5 = -7.5
Final Answer: -7.5
Note: this is the midpoint of -10 and -5 on the number line
I would assume it's the area of both the sidewalk and pool combined (79ft x 49ft = 3,871) minus the area of the pool alone (75ft x 45ft = 3,375). The area of the sidewalk would then be
496 ft
.
