We know that
Two angles are said to be co-terminal <span>if they have the same initial side and </span>
<span>the same terminal side.
</span>
(52π/5)-----> 10.4π<span>
so
</span>(52π/5)-5*2π------> (2π/5)
the answer is
the positive angle less than one revolution around the unit circle that is co-terminal with angle of 52π/5 is 2π/5
Answer:
just use a calculator and get 0.00166112956
The X-intercepts: (1,0),(-7,0)
Axis of symmetry: x= - 3
The vertex: (-3, -8)
The Y-intercept: (0, -7/2)
Concave up or down: concave up
Sketch: a poorly drawn picture on the bottom.
I hope that helps.
Answer:
- 6.04 km (per angle marks)
- 5.36 km (per side hash marks)
Step-by-step explanation:
Going by the angle indicators, the ratios of corresponding sides of the similar triangles are ...
x/2000 = 4200/3500
x = 2000·6/5 = 2400 . . . . yards
Then the distance of interest is ...
(2400 yd + 4200 yd)×(0.0009144 km/yd) = 6.6×.9144 km
= 6.03504 km ≈ 6.04 km
_____
Going by the red hash marks, the ratios of corresponding sides of the similar triangles are ...
x/2000 = 3500/4200
x = 2000·(5/6) = 5000/3 . . . . yards
Then the distance of interest is ...
(5000/3 + 4200) yd × 0.0009144 km/yd ≈ 5.36 km
_____
<em>Comment on the figure</em>
The usual geometry here is that the outside legs (opposite the vertical angles) are parallel, meaning that the angle indicators are the correct marks. It is possible, but unusual, for the red hash marks to be correct and the angle indicators to be mismarked. The red hash marks seem tentatively drawn, so seem like they're more likely to be the incorrect marks.
Answer:
Step-by-step explanation:
Since log is defined by all positive real numbers
therefore domain is all positive real number that is ( 0,∞)
Range is given by real numbers
inverse of the given function is (10^x)/7
Whose domain is all real numbers and range is all positive real number
And since we know that domain of function and range of its inverse
& range of a function and domain of its inverse is same
which we are getting in the problem
so answer is justified
