I don't know for sure, but I think there might be an error in the way you copied your function. In all of my dealings with these types of problems, I have learned the formula to be h(t) = 15 + 10t - 16t^2. Notice the plus in front of the "10t". This is due to the fact that if he pushes off of the overhang he would have an upward force of 10 feet per second as soon as his feet left the ground. The only thing pulling him back to Earth is gravity, modeled by the "-16t^2". This is derived from a bit of slightly advanced physics involving the gravitational constant, but let's work under my formula for a second...
Either way, we will wind up using the Quadratic Formula (or possibly factoring if the numbers are easy enough to work with). So let's start.
h(t)= 15 +10t - 16t^2
In order to use the QF or factoring I will need to make h(t)=0. Simply done by:
0= 15 +10t -16t^2
Looking at the numbers, I'd prefer to use the QF so here it is:
I know that my answer will need to be positive since you can't have a negative value when dealing with time, so I will eliminate the positive sign from the "+or-" part leaving me with:
And I know that a= -16, b= 10, and c=15. So all that's left to do is substitute and solve.
There's a decent amount of math that would be difficult and sloppy for me to do over the computer, but all you need to do is solve the rest of the equation and you would get your answer.
Exact answer:
or rounded answer: ≈1.31 seconds.
Hope this helps!
NoThisIsPatrick
Answer:
seven hundred ten and two hundred and thirty one 700,000+ 10,000+ 200+30+1
Step-by-step explanation:
Area of small circle = PI x r^2
r = 1/2 the diameter ( 1/2 * 3 = 1.5)
3.14 x 1.5^2 = 7.065 square cm
there are 2 small circles so total area for them = 7.065 * 2 = 14.13 square cm
the area for the large circle = PI x r^2
the radius of the large circle is 3 cm
3.14 x 3^2 = 28.26 square cm
the area of the shaded area is: 28.26 - 14.13 = 14.13 square cm, which can round down to 14 square cm.
Answer:
Step-by-step explanation:
To prove Δ ABC similar to ΔDBE we can consider
Segments AC and DE are parallel.
⇒ DE intersects AB and BC in same ratio.
AB is a transversal line passing AC and DE.
⇒∠BAC=∠BDE [corresponding angles]
Angle B is congruent to itself due to the reflexive property.
All of them are telling a relation of parts of ΔABC to ΔDBE.
The only option which is not used to prove that ΔABC is similar to ΔDBE is the first option ,"The sum of angles A and B are supplementary to angle C".
B : subtractions
because first of all is What is inside the arc