Y=18-7x/-4
gradient = 7/4
parallel lines, same gradient.
y--7=7/4(x-2)
y=7x-14-28/4
y= (7x + 14)/4
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 
:
it's C because the floor area is 120 square feet and the walls are 8 feet so if you multiply 120 × 8 you get 960.
Answer:
<u>Option A</u>
Step-by-step explanation:
To reflect line segment BC over line m, BB' will be perpendicular to the line m
and line m bisector of BB'.
<u>So, the correct answer is option A</u>
A) Line m is the perpendicular bisector of line segment BB' and the line segment CC'
<u>Option b is wrong</u> , it is impossible for the line B'C' to be perpendicular to line BC. B'C' is the image of BC.
<u>Both option c and d is wrong</u> because the perpendicular distance from b to the line m not equal to the perpendicular distance from c to the line m.
