Answer:(Fog)(-6)=29
Step-by-step explanation:
(fog)=-3(2x+2)-1
=-6x-6-1
=-6x-7
(fog)(-6)= -6(-6)-7
=36-7
=29
Answer:
The answer to your question is: 40°
Step-by-step explanation:
Due to the directions, we can know that KL is parallel to GH, then the
∠ FKL ≡ ∠ FGH = 40
Let
A (-4,35)
B (-3,20)
C (-2,10)
D (-1,5)
E (0,8)
F (1,10)
G (2,15)
H (3,30)
I (4,48)
using a graph tool
case a) y=2.23x²+1.62x+5.26
see the attached figure N 1
case b) y=4.63x²+0.84x+4.14
see the attached figure N 2
case c) y=0.98x²+3.06x+6.02
see the attached figure N 3
case d) y=3.72x²+2.94x+3.21
see the attached figure N 4
using a <span>Quadratic Regression Calculator
</span>the <span>Best Fit Second-Degree Quadratic Regression is
2.228x</span>²+1.617x+5.255
see the attached figure N 5
therefore
the answer iscase a) y=2.23x²+1.62x+5.26
Have you heard the expressions "shooting fish in a barrel" or
"taking candy from a baby" ?
I've decided to take your points and try not to feel guilty about it.
A). That's it ! That's the equation, for ANY basketball on Earth.
You wrote it right there in your question.
In that equation that you wrote, the ' r ' is the initial velocity.
You said that Adam tossed it straight up, and it was going 10 meters per second
when it left his hand.
So ' r ' in the equation is +10.
The equation is:
<em>d = 10t - 5t² </em> .
It tells you how high the ball is above its tossing height after any number of seconds.
The ' t ' in the equation is the number of seconds. Any letter could have been used,
but ' t ' was cleverly selected because it stands for 'time'.
B). You want to know where it is after 0.5 second ? All you have to do is put '0.5'
into the equation wherever there's a ' t '. Do you really need somebody to do that
for you ?
Well, OK. You're being so overly generous with your points . . .
Distance = 10 t - 5 t²
Distance = 10 (0.5) - 5 (0.5)²
Distance = 5 - 5(0.25)
Distance = 5 - 1.25
<em>Distance = 3.75 meters </em>
<u>positive</u> 3.75 means <u>above</u> Adam's hand.
A thinking exercise:
-- He tossed the ball upwards at 10 meters per second.
-- Why is it not 5 meters above his hands after 0.5 second?