Answer: -8x^3y^2 (choice B)
The coefficients multiply to 4 times -2 = -8
the x terms multiply to x^2 times x = x^(2+1) = x^3
The y terms multiply to y times y = y^(1+1) = y^2
So that's how I ended up with -8x^3y^2
H^2=13^2-5^2
h^2=169-25
h^2=144
h=V144
h=12
A=b*h/2=10*h/2
A=10*12/2=120/2=60 cm^2
I don’t see a figure but I’ll be happy to help
For the time period of the projectile at least 192 feet above the ground is between 2 second and 6 second
Since s=128t-16t2, we want to know the 2 times where 128t-16t2=192, as those two values will be the begin and end times of the interval in question. We can rewrite that equation in standard quadratic equation form:
16t2-128t+192=0 or, simplified, 8t2-64t+96=0
or t2-8t+12=0
since it is now in quadratic form, we can solve using the quadratic formula:
t = 2 and 6
So the time period from approximately 2 second and 6 second has the projectile above 192 feet.
Learn more about projectile here:
brainly.com/question/1659208
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