Answer:
p-q= -4
p=q-4
x^2 + px + q=0
=> x^2 + (q-4)x + q = 0
d= b^2-4ac
=(q-4)^2 - 4(1)(q)
= q^2 + 16 -8q - 4q
=q^2 -12q+16
given that D=16
q^2 -12q+16 = 16
=> q^2 -12q = 0
=> q^2 = 12q
=> q = 12q/q
=> q= 12
Therefore the value of q is 12 and the value of p is q-4 = 12-4= 8
Hope you understood............
Answer:
Yes
Step-by-step explanation:
Yes it is a function, because graph is intersecting y - axis only once.
Answer:
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
Step-by-step explanation:
If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -
x = ( 30 cos 20° )( time ),
y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2
To determine " ( 30 cos 20° )( time ) " you would do the following calculations -
( x = 30 * 0.93... = ( About ) 28.01t
This represents our horizontal distance, respectively the vertical distance should be the following -
y = 30 * 0.34 - 4.9t^2,
( y = ( About ) 10.26t - 4.9t^2 + 2
In other words, our solution should be,
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
<u><em>These are are parametric equations</em></u>
