Please help! and explain thank you!
2 answers:
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Ooh this is a good one!
Firstly, let's find RT so that we can solve the top triangle.
We could actually use 3 methods to find this side: 1) trig to find RT
2) trig to find ST, then Pythagorean Theorem to determine RT
3) use the special 30-60-90 right triangle rule ... hmm, I'm going with the last because it's easiest ;)
For 30-60-90 triangles, the sides are always as follows: smallest side (opposite the 30) is s, the middle side is sSR3, and the hypotenuse is 2×s
Thus, since we have the side opposite 60, it is = sSR3, so:
So now, the hypotenuse (RT) = 2s = 2(2) = 4
Secondly, we are looking at the other type of special right triangle, the 45-45-90:
The side opposite the 45s are s (remember side lengths are equal whose opposite angles are equal); and the hypotenuse is always
we only need QR, which is s, which is also = RT
So the answer is:
QR = x = 4
x = 4 IS THE FINAL ANSWER!!!
[ i hope by my explanations you understand how better to do special right triangles in the future] :-D
Firstly, let's find RT so that we can solve the top triangle.
We could actually use 3 methods to find this side: 1) trig to find RT
2) trig to find ST, then Pythagorean Theorem to determine RT
3) use the special 30-60-90 right triangle rule ... hmm, I'm going with the last because it's easiest ;)
For 30-60-90 triangles, the sides are always as follows: smallest side (opposite the 30) is s, the middle side is sSR3, and the hypotenuse is 2×s
Thus, since we have the side opposite 60, it is = sSR3, so:
So now, the hypotenuse (RT) = 2s = 2(2) = 4
Secondly, we are looking at the other type of special right triangle, the 45-45-90:
The side opposite the 45s are s (remember side lengths are equal whose opposite angles are equal); and the hypotenuse is always
we only need QR, which is s, which is also = RT
So the answer is:
QR = x = 4
x = 4 IS THE FINAL ANSWER!!!
[ i hope by my explanations you understand how better to do special right triangles in the future] :-D
Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this
So, i integrate this,
I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
So, i integrate this,
I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4