<h3>
Answer: A. 18*sqrt(3)</h3>
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Explanation:
We'll need the tangent rule
tan(angle) = opposite/adjacent
tan(R) = TH/HR
tan(30) = TH/54
sqrt(3)/3 = TH/54 ... use the unit circle
54*sqrt(3)/3 = TH .... multiply both sides by 54
(54/3)*sqrt(3) = TH
18*sqrt(3) = TH
TH = 18*sqrt(3) which points to <u>choice A</u> as the final answer
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An alternative method:
Triangle THR is a 30-60-90 triangle.
Let x be the measure of side TH. This side is opposite the smallest angle R = 30, so we consider this the short leg.
The hypotenuse is twice as long as x, so TR = 2x. This only applies to 30-60-90 triangles.
Now use the pythagorean theorem
a^2 + b^2 = c^2
(TH)^2 + (HR)^2 = (TR)^2
(x)^2 + (54)^2 = (2x)^2
x^2 + 2916 = 4x^2
2916 = 4x^2 - x^2
3x^2 = 2916
x^2 = 2916/3
x^2 = 972
x = sqrt(972)
x = sqrt(324*3)
x = sqrt(324)*sqrt(3)
x = 18*sqrt(3) which is the length of TH.
A slightly similar idea is to use the fact that if y is the long leg and x is the short leg, then y = x*sqrt(3). Plug in y = 54 and isolate x and you should get x = 18*sqrt(3). Again, this trick only works for 30-60-90 triangles.
So firstly, I'm going to be converting the mixed number into an improper fraction. To do this, multiply the whole number by the denominator and then add that product with the numerator and that will be your new numerator. In this case:
3 × 8 = 24, 24 + 3 = 27.
Now since they have the same denominator, you can subtract the numerators.
Now simplify the number as such:
<u>In short, your answer is -17/4, or -4 1/4.</u>
The speed of one bicyclist was 14.5mph, speed of the other bicyclist was 17.5mph.
Let the speed of one bicyclist=x mph
Let the speed of the other bicyclist=(x+3) mph
Hence:
Speed of one bicyclist:
3x+3(x+3)+2=98
3x+3x+9+2=98
6x=87
Divide both side by 6x
x=87/6
x=14.5 mph
Speed of the other bicyclist:
x+3 mph
14.5 mph+3 mph
=17.5 mph
Inconclusion the speed of one bicyclist was 14.5mph, speed of the other bicyclist was 17.5mph.
Learn more here:
brainly.com/question/18839894
a. Assume a is even, so a = 2k for some integer k. Now let a and b be integers such that a divides b and a + b is odd.
Since a divides b, b = an for integer n, and in turn b = 2nk, which means <u>b is even</u> and hence a + b is also even. But this contradicts our initial assumption, so a must be odd.
b. Let n be even, so that n = 2k for some integer k. Then
n² = (2k)² = 4k²
so that n² ≡ 0 (mod 4).
Now let n be odd, so n = 2k + 1 for integer k. Then
n² = (2k + 1)² = 4k² + 4k + 1
so that n² ≡ 1 (mod 4).
Therefore n² is never congruent to 2 (mod 4).