Answer:
0.5
Step-by-step explanation:
the first significant figure is the non zero
Answer: its b
Step-by-step explanation:
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 
:
The general equation for a circle with center (a,b) and squared radius k is
Here we have
and the constant is fixed by substituting in the point we know (x,y)=(3,-5)
Answer:
Answer for problem 1: slope = 2, y intercept = 3
Answer for problem 2: slope = 3/4, y intercept = -2
Explanation:
Both equations are in the form y = mx+b. This is known as slope intercept form. This is because we can read off the values of the slope and y intercept very quickly. We have y = 2x+3 match up with y = mx+b. The m is the slope and b is the y intercept. So m = 2 and b = 3 for this equation. A similar situation happens with the other equation as well. It might help to rewrite the second equation into y = (3/4)x + (-2).
