The slope for the first one would be:
y2-y1/x2-x1, so replace those with the coordinates and you'll get:
-3-1/-7-(-7) => -4/0 so I guess the slope is zero
the slope for the second one would be:
-3-(-3)/5-(-4)=> 0/9 I think this one would be undefined.
Check to make sure, though!
The computation shows that the placw on the hill where the cannonball land is 3.75m.
<h3>How to illustrate the information?</h3>
To find where on the hill the cannonball lands
So 0.15x = 2 + 0.12x - 0.002x²
Taking the LHS expression to the right and rearranging we have:
-0.002x² + 0.12x -.0.15x + 2 = 0.
So we have -0.002x²- 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x² + 0.03x -2 = 0.
This is a quadratic equation with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25.
The second solution y = 0.15 * 25 = 3.75
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Complete question:
The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?
Twice the difference of a number and 3 is at most 27
2*(n-3) <=27
to solve
distribute
2n -6 <=27
add 6 to each side
2n <=33
divide by 2
n <=33/2
n<= 16 1/2
Answer:
No.
Step-by-step explanation:
Y=2x goes up twice and to the right once, it passes through the origin, therefore it won't touch or go pass by point (2,1)
1200 +29%= (1200+240)=1440
1440-300=1140
1140÷10=114
so the answer is £114 per month for 10 months
