<span>This question requests to find the inverse function of sine function. The inverse function of sine function is called arc sine. This is, the arc length whose sine is the value searched. Then you need to find the arc sin (0.6123). the result may be in radians or degrees. Only in some particular cases you can tell the arc sine without a calculator or table, it is when the sine value is a notable number (e.g. 0, 1, -1, or some others) In this case you need to use a calculator or table. With a scientific calculator make sure the mode is in degrees and then enter the number 0.6123. With that numbered entered now press the key sin^-1, which is the key for the inverse of the sine function. I did it and got 37.76 degrees. Rounded to the nearest whole degree that is 38. Then, the answer is 38 degrees. </span>
Answer: 200 in
Step-by-step explanation:
Try to split up the face into pieces. You can make the first piece the square on the top. Only, you don't know the value of the bottom side, but you can see that it is 10 (if you look at the top at the back). So, if you do your multiplication, 10 x 10 = 100. So, that is the area of the first square. The next square also has a width and height of 10, (10 on the bottom, and follow the side to the other 10). Do the multiplication again (same as above) and get 100. Now that you have the area of the two pieces, you add then together: 100+100=200
Please feel free to let me know if you have any questions!
-10 and 2
-10 x 2 = -20
-10 + 2 = -8
Answer:
Triangle PQR is drawn on a coordinate plane with vertices P (-6, 0), Q (-3, 4), and R(2, 4). What is the length of PR? Is APQRisosceles (two sides of equal length)? Use the drop-down menus to answer the questions. The length of PR is approximately Choose. units. Triangle PQR Choose. an isosceles triangle because Choose.. v are the same -2-1 Choose.. length. IS is not
BG ≅ AG
BG is the perpendicular to the side of the triangle while AG is the angle bisector , So BG cannot equal AG , So BG cannot be congruent to AG. Hence first is false.
DG ≅ FG
DG And FG both are the perpendicular to the sides from the incentre of the circle , Hence DG and FG are congruent , So second statement is true.
DG ≅ BG
Again DG and BG both are the perpendicular to the sides from the incentre of the circle , Hence DG and BG are congruent , So third statement is true.
GE bisects ∠DEF
As said in the question GE is the angle bisector , So yes GE bisects ∠DEF.
This Statement is true.
GA bisects ∠BAF
Again As said in the question GA is the angle bisector , So yes GA bisects ∠BAF.
Hence 2nd, 3rd , 4th , and 5th options are correct.
