Answer:
C
Step-by-step explanation:
There’s nothing hard with this
You need to know special right triangles
This is a 30,60, 90 triangle
We automatically know becuase theres. Right angle (90) and a 60 degree angle
now the formula is the smallest side is (n) {I’m using N becuase theres already an X) or in this case, 6
The hypotenuse, the one directly above the 2nd biggest side, or the diagnoal side is 2n or (12). Now we know what X is in this situation which is 12. So that narrows it to 2 answers
Now the side on the bottom of the hypotenuse, the second biggest side is used in the formula N. So we know what N is in the beginning, 6 so we just plug that in and well get C. Attached is a photo on q 30, 60, 90 special right triangle
Time while going up = time while going down.
One-way time = 0.47 / 2 = 0.235 s
While going up,
s = ut + 1/2 at²; where u is the initial velocity upwards and a is the acceleration due to gravity downwards.
2.14 = u(0.235) + 1/2 x -9.81 x (0.235)²
u = 10.2 m/s
Max height attained can be obtained by:
2as = v² - u²; where v is 0
s = -(10.2)² / -9.81 x 2
s = 5.30 m
The height above the window is:
5.30 - 2.14
= 3.16 m
Step-by-step explanation:
yes its ans is (3,3) fast you have see in x axis then y axis .so according to this we write 3 which lies in x- axis and then another 3 which lies in y-axis.
Answer:
. (In this problem we prove a fact that you demonstrated experimentally in Problem1 of the fourth assignment.) LetABCDbe a quadrilateral. LetM, N, P,andQbe the midpoints of the sides. Prove the area ofMNPQis one half the area ofABCD.4. (See Figure 1.) Give the proof of Theorem 24 for Case (iii). Given:MandNarethe midpoints ofABandAC,MX⊥AB,NX⊥AC, andXis onBC. To prove:Xis on the perpendicular bisector ofBC.XNCMABFigure 1
5. (See Figure 2). Prove Case (ii) of Theorem 28. Given:A0,B0andC0are collinear.To prove:A0BA0CB0CB0AC0AC0B= 1.C'A'B'ABCFigure 26. (See Figure 3.)Given:6A=6B,AD=BE,6ADG=6BEF.To prove:6CFE=6CGD.FGECDBAFigure 37. Suppose that you have a computer program which can perform the following func-tions:
(a) It can draw points, and draw line segments connecting two points.(b) Given a pointOand a line segmentAB, it can construct the circle with centerOand radius equal to the length ofAB.(c) Given a line segmentAB, it can find the midpoint.(d) Given a lineland a pointP(not necessarily lying onl), it can construct theline throughP
Step-by-step explanation: