Answer:
both are c
Step-by-step explanation:
<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.
Answer:
If the line RS has been rotated 90 degrees, then VU will be perpendicular to RS and the two slopes must be opposite and reciprocal, i.e. product of the two slopes will equal -1.
As a verification, we find the locations of V and U from rotations of R & S.
(actually, the triangle had been rotated -90°, 90 ° clockwise)
Step-by-step explanation:
Slope RS, m1:
Slope VU, m2
Hence m1*m2=1*-1=-1, meaning that m1 and m2 are opposite (in sign) and are reciprocal to each other, as expected
Answer:
Domain = {x : x ≠ 4 , -4} or (-∞ , -4) ∪ (-4 , 4) ∪ (4 , ∞)
Step-by-step explanation:
<u>TO FIND :-</u>
- Domain of
<u>SOLUTION :-</u>
Domain of a function is a value for which the function is valid.
The function is valid until the denominator is 0.
So make sure that the denominator must not be 0.
Find the values of x for which the denominator becomes 0. To find it , you'll have to solve the above inequality.
We can say that <u>4 & -4 can't be domains</u> because these values will make the function undefined.
Now try putting values of x such that -4 < x < 4. You'll observe that the function will be valid for all those values of x between -4 & 4.
<u>CONCLUSION :-</u>
The function will be valid for any value of 'x' except 4 & -4. So in :-
Interval notation , it can be written as → (-∞ , -4) ∪ (-4 , 4) ∪ (4 , ∞)
Set builder notation , it can be written as → {x : x ≠ 4 , -4}
Well in my answer i got 5.06 i rounded to the nearest hundred instead of putting all of the numbers behind the 6.
