Answer: y = 4 and x = 1
Step-by-step explanation:
when y is 2, x is 0.5
when y is 4, x is 1
.(+ 2) .(+ 0.5)
.(+ 2) .(+ 0.5)
.(+ 2) .(+ 0.5)
.(+ 2) .(+ 0.5)
the rule is that y is 4 times larger than x
p(x)= x-2
g(x)= 2x^3 + 3x^2 - 11x - 6
first we have to find the zero of the polynomial of x-2
p(x)= x-2 = 0
x=2
therefore,
p(x)= 2x^3 + 3x^2 - 11x - 6
p(2)= 2*2^3 + 3*2^2 - 11*2 - 6
= 2*8 + 3*4 - 11*2 - 6
= 16 + 12 - 22 - 6
= 28-28
= 0
Hope it helped u, ^_^.
Answer:
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:
(x-4)=0 ===> x=4,
so that now you have to plot 2 functions!
For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:
|x-4| = -(x-4)=4-x
Then:
for x<=4, y = -x+4-7 = -x-3
for x=>4, (x-4) is positive, so no changes:
y= x-4-7 = x-11,
Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11
In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:
x=4 and x= 3 for y=-x-3
And just x=5 for y=x-11
The reason is that the absolute value is continuous, so x=4 works for both:
x=4===> y=-4-3 = -7
x==4 ====> y = 4-11=-7!
abs() usually have a cusp int he point where it is =0
Step-by-step explanation:
Answer:
Step-by-step explanation:
<h2>A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second.
</h2><h2>y=-16x^2+129x+119
</h2><h2>y=−16x </h2><h2>2
</h2>
+129x+119
