Y+9=-3x can be turned into y=-3x-9 by subtracting 9 on both sides of the equation.
When you graph y=-3x-9 you find that the x-intercept is -9
Answer:
<h2>See the explanation.</h2>
Step-by-step explanation:
Sin2α = 2SinαCosα.
Cos2α = (Cosα)^2 - (Sinα)^2.
Sin4α can be written as Sin2(2α).
Similarly Cos4α can be written as Cos2(2α).
It is given that tanα=3.
Sinα(Sinα) = [], the values either be both positive or both negative, as the value of tanα is positive only on first and third quadrant.
Sin4α = .
Cos4α =
Answer:
Mean = 0
Variance = 4/3
Standard Deviation √4/3
a= 0.9
Step-by-step explanation:
If X has a uniform distribution over [a,b] then its Mean is a+b/2 and variance is (b-a)²/12
Here a= -2 and b= 2
Now finding the mean = a+b/2=-2+2/2= 0
Variance = (b-a)²/12=( 2-(-2))²/12= 4²/12= 16/12= 4/3
Standard Deviation = √Variance= √4/3
b) = \int\limits^a_a {\frac{1}{a- (-a)} } \, dx
=1/2a[x]^a_-a= 2a/2a= 1 (applying the limits to the function)
P(−a<X<a) ==1/2 * 2a= a (applying the limits to the function)
P(−a<X<a)= 0.9
a= 0.9
In the given question the limits are -a to a . When we apply these in the above instead of [a,b] we get the above answer.
B. The answer is b hehehehhehsheus
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