Answer:
Yes, there are infinite triangles with the same three angles but different side lengths
Step-by-step explanation:
we know that
If two triangles are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent
therefore
There are infinite triangles with the same three angles but different side lengths
1x24, 2x12, 3x8 are 3 ways that get 24
The area of D is given by:
The average value of f over D is given by:
![\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Cfrac%7B343%7D%7B3%7D%20%7D%20%20%5Cint%5Climits%5E7_0%20%20%5Cint%5Climits%5E%7Bx%5E2%7D_0%20%7B4x%5Csin%28y%29%7D%20%5C%2C%20dydx%20%20%3D%20-%5Cfrac%7B3%7D%7B343%7D%20%20%5Cint%5Climits%5E7_0%20%7B4x%5Ccos%28x%5E2%29%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D-%5Cfrac%7B3%7D%7B343%7D%20%5Cint%5Climits%5E%7B49%7D_0%20%7B2%5Ccos%28t%29%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%20%5Cleft%5B%5Csin%28t%29%5Cright%5D_0%5E%7B49%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%5Csin49)
Step-by-step explanation:
Use arc length:
s = rθ
where r is the radius and θ is the angle in radians.
a) First, convert the angle to radians (1 degree = 60 minutes).
77 ⁵⁰/₆₀° × (π radian / 180°) = 1.358 radians
s = rθ
s = (9.67 in) (1.358 rad)
s = 13.1 in
b) s = rθ
4 in = (9.67 in) θ
θ = 0.414 radians
θ = 23.7°
θ = 23° 42'
Addition is the correct answer.
x^2 is the part where you get the second degree term. If you add x^2+x^2 you get 2x^2. If you subtract x^2-x^2 you get 0. If you multiply x^2*x^2 you get x^4, which is a fourth degree term
