Answer:
Step-by-step explanation:
t - 1 - t^3 - 4t
-t^3 - 3t
Answer:
false
Step-by-step explanation:
thats not fair for the other team
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:
Part 1) The base is 
and the height is 
Part 2) The base is 
and the height is 
Step-by-step explanation:
case 1) Right isosceles triangle of the left
Let
x------> the base of the rectangle
y----> the height of the rectangle
Remember that
In a right isosceles triangle the lengths of the legs of the triangle is the same
----> equation A
-----> equation B
substitute equation B in the equation A
Find the value of x
case 2) Right isosceles triangle of the right
Let
x------> the base of the rectangle
y----> the height of the rectangle
Remember that
In a right isosceles triangle the lengths of the legs of the triangle is the same
----> equation A
-----> equation B
substitute equation B in the equation A
Find the value of y
Tan=opp/adj
Tan37=opp/13
X=13tan37
