Answer:
False
Step-by-step explanation:
I think the answer would be 343?
For C highest to lowest
4/5, 3/4, 7/10, 1/2
For D
3/4, 17/24, 2/3, 7/12, 1/6
Answer:
Step-by-step explanation:
The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be 
. The magnitude of 
will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is 
, we have:
(Recall that 
)
Now that we've found the vertical component of the velocity and launch, we can use kinematics equation 
to solve this problem, where 
is final and initial velocity, respectively, 
is acceleration, and 
is distance travelled. The only acceleration is acceleration due to gravity, which is approximately 
. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.
What we know:
Solving for 
:
Answer:
8. x = 16
9. x = 10
14.
m ∠RSU = 130°
m ∠UST = 50°
15.
m ∠RSU = 124°
m ∠UST = 56°
Step-by-step explanation:
8.
Given ∠DEF is bisected by EG. That is , ∠DEG = ∠GEF
That is , (x + 15)° = 31°
x = 31 - 15 = 16
9.
Given ∠DEF is bisected by EG. That is , ∠DEG = ∠GEF
That is ,
(6x - 4)° = 56°
6x = 56 + 4
6x = 60
x = 10
14.
13x + 5x = 180° [straight line angles ]
18x = 180
x = 10
m ∠RSU = 130°
m ∠UST = 50°
15.
4x + 12 + 2x = 180° [ straight line angles]
6x = 180 - 12
6x = 168
x = 28
m ∠RSU = 4(28) + 12 = 112 + 12 = 124°
m ∠UST = 2(28) = 56°
