The answer is the 2nd one.
Hope it helped!
May I presume that the question is "How long does it take
the object to reach its maximum height ?" ?
If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
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If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now, and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The picture in the attached figure
we know that
The Intersecting Secants Theorem, states that
''If two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment''
so
in this problem
(x+DC)*DC=(AB+BC)*BC
AB=42
BC=18
CD=4
(x+4)*4=(42+18)*18
4x+16=1080
4x=1080-16
x=1064/4
x=266
the answer isx=266
Answer:
B = 68° and C = 22°
Step-by-step explanation:
Let angle B's measure be B and angle C's measure be C.
Also note that complementary means that they sum to 90.
Now:
<u>The m∠B is two more than three times the measure of ∠C:</u>
we can write:
B = 3C + 2
<u>If ∠B and ∠C are complementary angles:</u>
B + C = 90
Putting 1st equation in 2nd, we get:
B + C = 90
3C + 2 + C = 90
4C = 90 - 2
4C = 88
C = 88/4 = 22
Now B = 3C + 2, so
B = 3(22) + 2 = 68
Hence, B = 68 degrees and C = 22 degrees