If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±<span>√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.
The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151
Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.
Hope this makes sense.
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Answer:
y=2x+1
Step-by-step explanation:
First you have to find the slope:
5-1/2 so 4/2 or 2
Then you write it in point-slope: (I used the point (0,1))
y-1=2x
Simplify than:
y=2x+1
Answer:
Banners: 4
Paper lanters: 3
Bags of Balloons: 6
Garlands: 11
Step-by-step explanation:
Banner: 4*5.25= $21
Paper lantern: 6.50*x=?
Balloons: 5*x=?
Garland: 3.50*x=?
Paper lantern: 145.50/6.50= $22.3
22.3/6.50= 3.4 round to 3
3 paper lanterns
Balloons: 145.50/5= $29.1
29.1/5= 5.8 round to 6
6 packs of balloons
Garland: 145.50/3.50= $41.5
41.5/10= 4.15
4.15+7=11.15 round to 11
11 garlands
Answer:
R is a function.
Step-by-step explanation:
