The test lasts for 1 hour and 35 minutes. If they finish the test at 1:20, what time did they start the test
1 answer:
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Answer:
- ∠R = 56°
- ∠Q = 90°
- ∠S = 34°
Step-by-step explanation:
The given triangle is a right angled triangle.
So, the angles in the triangle are :
- 90°
- (2x + 38)°
- (5x - 11)°
Solving according to <u>angle sum property</u>,
Sum of all angles in a triangle is 180°
90° + (2x + 38)° + (5x - 11)° = 180°
117° + 7x = 180°
7x = 180° - 117°
7x = 63°
x = 9
Angles =
2(9) + 38
56°
5(9) - 11
34°
- ∠R = 56°
- ∠Q = 90°
- ∠S = 34°
The angles are 56°, 90° and 34°.
Answer:
The Taylor series of f(x) around the point a, can be written as:
Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as: