How many degrees did the clock turn from 2:30 p.m. to 5:15 p.m? How many degrees did the clock turn from 2:30 p.m. to 5:15 p.m?

How many degrees did the clock turn from 2:30 p.m. to 5:15 p.m? How many degrees did the clock turn from 2:30 p.m. to 5:15 p.m?

The hour hand rotates 360 degrees every day, 360 / 12 = 30 degrees per hour, 30 / 60 = 0.5 degrees per minute
2: 30 to 3:15, 45 minutes, 3:15 to 5:15, 2 hours, 2 * 60 + 45 = 165 minutes
0.5 * 165 = 82.5 degrees
A: turn the clock 82.5 degrees

From 2:15 to 2:35, the minute hand of the clock is gone______ Degrees, the clock is gone______ Degree

∵ the hour hand turns 0.5 ° per minute on the clock face, and the minute hand turns 6 ° per minute. After 20 minutes from 2:15 to 2:35, the minute hand of the clock moves 6 ° x 20 = 120 ° and the hour hand moves 0.5 ° x 20 = 10 °

The clock goes from 2:30 to 2:55. How many degrees does the hour hand go and how many degrees does the minute hand go?

When the clock goes from 2:30 to 2:55, both the hour hand and the minute hand go for 25 minutes

(2 / 17-4 / 13) - (2 / 17-3 / 7) - (9 / 13-4 / 7) - quick solution~~

(2 / 17-4 / 13) - (2 / 17-3 / 7) - (9 / 13-4 / 7)|
=2/17-4/13-2/17+3/7-9/13+4/7
=(2/17-2/17)-(4/13+9/13)+(3/7+4/7)
=0-1+1
=0

As shown in the figure, in a closed container, a certain amount of metal sodium and a certain amount of mercuric oxide are filled on one side, and the two parts of the container are heated at the same time. After both sodium and mercuric oxide are completely reacted, the original temperature is restored, and the composition of the air in the container remains unchanged. Then the ratio of the amount of sodium and mercuric oxide is ()
A. 1：1B. 2：1C. 23：216.6D. 216.6：23

After the complete reaction, it returns to the original temperature, and the composition of the air in the container remains unchanged, which means that the replacement reaction of sodium and HGO produces sodium peroxide and Hg. The reaction equation is 2Na + 2hgo = Na2O2 + 2hg. According to the equation, the ratio of Na and HGO is 1:1, so a

If a triangle and a parallelogram are equal in height, the difference between their areas is 30m2, and the area of the triangle is (?) m2

If a triangle and a parallelogram are equal in height, the area of the parallelogram is twice that of the triangle, and the difference of their areas is the area of a triangle, so the area of the triangle is 30m & sup2

Simple operation of 38 / 25

=（38×4)/(25×4)
=152/100
=1.52

As shown in the figure, we know that the perimeter of an isosceles triangle is 16, and the height of its bottom edge is 4

Let BD = x, then AB = 8-x. from Pythagorean theorem, we can get AB2 = BD2 + ad2, that is, (8-x) 2 = x2 + 42, ∧ x = 3, ∧ AB = AC = 5, BC = 6

In the isosceles trapezoid ABCD, ad ‖ BC, ad = DC = AB, BD = BC, find the degree of ∠ a

∫ isosceles trapezoid ABCD ≠ ∠ ABC ＝ C ∫ ad = ab ≠ ∠ abd ＝ ADB ∫ AD / / BC ≠ ADB ＝ CBD ≠ ∠ abd ＝ CBD ≠ ABC / 2 ∫ ADB ＝ ABC / 2 ∫ ADB ＝ C / 2 ∫ BD = BC ∫ BDC ∫ AD / / BC ∫ C + ∠ BDC +} ADB = 180 ∫ C + ∠ C +} C / 2 = 180 ∫

It is known that one root of the square of the equation x-4x-p + 2p + 2 = 0 is Q, then q =?

delta=16-4(-p^2+2p+2)=4[p^2-2p+2]=4[(p-1)^2+1]>0
q=2±√delta/2=2±√[(p-1)^2+1]
Is there something missing from the title?