Answer:
y=-8/3x-11 is the equation of the line and the value of y
Step-by-step explanation:
Use point slope form m = y-y1/x-x1, or y-y1=m(x-x1)
Find the slope from the two points (-6,5) and (-3,-3)
m = y2-y1/x2-x1 = -3-5/-3+6 = -8/3
use one point say (-6,5) that passes through the line and the slope to form the equation y-5=-8/3(x+6) rearrange y= -8/3(x+6)+5
y=-8/3x - 48/3 +5 = -8/3x-16+5 = -8/3x-11
Answer:
<B = 47°
<C = 28°
b = AC = 28.0
Step-by-step explanation:
Given:
∆ABC
AB = c = 18
BC = a = 37
<A = 105°
Required:
Length of AC = b
measure of angle B and angle C
SOLUTION:
==>Use the sine rule, sin A/a = sinC/c to find the angle of C:
SinA = sin(105) = 0.9659
a = 37
sinC = ?
c = 18
0.9659/37 = sinC/18
Cross multiply
0.9659*18 = 37*sinC
17.3862 = 37*sinC
Divide both sides by 37
17.3862/37 = sinC
0.4699 = sinC
sinC = 0.4699
C = Sin-¹(0.4699)
C = 28.0° (nearest tenth)
==>Find angle B using sum of angles in a triangle:
Angle B = 180 - (105+28)
Angle B = 180 - 133
Angle B = 47°
==>Find length of b using sine rule, b/sinB = c/sinC:
SinC = sin(28) = 0.4695
SinB = sin(47) = 0.7314
c = 18
b = ?
b/0.7314 = 18/0.4695
Cross multiply
b*0.4695 = 18*0.7314
b*0.4695 = 13.1652
Divide both sides by 0.4695
b = 13.1652/0.4695
b = 28.0 (nearest tenth)
Answer:
{1,4}
Step-by-step explanation:
A={3,6,9}
B={2,3,5,7}
C={1,2,3,4,5}
AUB={2,3,5,6,7,9}
(AUB)'={1,4,8,10}
(AUB)'⋂C={1,4}
The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.
Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s
The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.
Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s
When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.
Answer: 7 s