A.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.
h'(t) = -32t + 16
When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.
-32t + 16 = 0
-32t = -16
t = 0.5 seconds
b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.
h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet
If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.
c.) We know that the horse is in the air whenever h(t) is greater than 0.
-16t^2 + 16t = 0
-16t(t-1)=0
t = 0 and 1
So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.
Recall that A = 1/2bh.
We are given that h = 4+2b
So, putting it all together:
168 = 1/2 b(4+2b)
168 = 1/2(4b + 2b^2)
168 = 2b + b^2
b^2 + 2b - 168 = 0.
Something that multiplies to -168 and adds to 2? There's a trick to this.
Notice 13^2 = 169. So, it's more than likely in the middle of the two numbers we're trying to find. So let's try 12 and 14. Yep. 12 x 14 = 168. So this factors into (b+14)(b-12) So b = -14 or b =12. Is it possible to have a negative length on a base? No. So 12 must be our answer.
Let's check this. If 12 is our base, then according to our problem, 2*12 + 4 would be our height... or 28. so what is 12 * 28 /2?
196. Check.
Hope this helped!
120 different 3digit numbers
<span>3x + y = 9 (I)
</span><span>y = –4x + 10 (II)
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Pass the incognito "4x" to the first term, changing the signal when changing sides.
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simplify by (-1)
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</span></span><span>Substitute in equation (I) to find the value of "Y".
</span>3x + y = 9 (I)
3*(1) + y = 9
3 + y = 9
y = 9 - 3

Answer:
