Set h to 640 and solve for t:
640 = -490t^2 + 1120t
Subtract 640 from both sides:
-490t^2 + 1120t - 640 = 0
The formula to solve a quadratic equation is:
x = -b -/+ sqrtroot (b^2-4ac)/(2a) where a = -490, b = 1120 and c = -640
Solve:
x = -1120 -/+ sqrtroot (1120^2-4(-490)(-640) )/ 2(-490)
x = 8/7 = 1.1428 = 1.14
Time was 1.14 seconds
Solve for x for x: x+2y=2
x+2y +-2y=2+-2y (add -2y on both sides.)
x=-2y+2
Substitute : -2y+2 for x in
4x-y=4:
4(-2y+2)-y=4
-9y+8=4 Simplify both sides of the equation.
-9y+8+-8=4+-8 Add -8 to both sides.
-9y=-4
-9/-9=-4-9 Divide both sides by -9
y=4/9
Substitute 4/9 for y in
x=-2y+2
x=-2(4/9)+2
x=10/9
y=4/9,x=10/9
Answer:
Part A: Angle R is not a right angle.
Part B; Angle GRT' is a right angle.
Step-by-step explanation:
Part A:
From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).
Slope formula

The product of slopes of two perpendicular lines is -1.
Slope of GR is

Slope of RT is

Product of slopes of GR and RT is

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.
Part B:
If vertex T translated by rule

Then the coordinates of T' are


Slope of RT' is

Product of slopes of GR and RT' is

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.
Answer:
4x - 5y + 30 = 0
Step-by-step explanation:
When a slope of a line and a point passing through it is given then we use slope - one point form to determine the equation of the line.
It is given by:

where
is the slope of the line and
is the point passing through it.
Here
and
.
Substituting in the equation we get



It is A because the radius is 10 which means that if x=0 then y must equal 10. (0,5) would be inside of the circle. If this doesn't make sense graph the circle and the points and you'll see what I mean.