C. No, this is not a valid inference because she asked only 35 families
The length of ND = 3 units
The length of N'D' = 6
6 / 3 = 2
The scale factor is 2
<h3>
Answer: 3 m/s^2</h3>
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According to Newton's Second Law, we know that
F = m*a
where F is the force applied, m is the mass and 'a' is the acceleration.
We see that this is a direct variation equation for F and a, such that m is the constant of variation. It's similar to how y = kx is also a direct variation equation.
Plug in F = 35 and a = 5 to find m
F = ma
35 = m*5
35/5 = m
7 = m
m = 7
The object has a mass of 7 kg
Our equation F = ma updates to F = 7a
Now plug in the force F = 21 to find 'a'
F = 7a
21 = 7a
21/7 = a
3 = a
a = 3
The acceleration will be 3 m/s^2
Notice how a smaller force applied means that the acceleration has also gone down as well.
Answer:
It is not a factor
Step-by-step explanation:
It is because p(x)=-12 at x=3
1. Use a straightedge to draw line m and label a point on the line as point F
2. Construct a line perpendicular to line m through point F. Label a point on this line as point G.
3. With the compass open to the desired side length of the square, place the compass point on point F and draw an arc on line m and an arc on FG←→ . Label the points of intersection as points H and K.
4. Without changing the compass width, place the compass point on point H and draw an arc in the interior of ∠HFK.
5. Keeping the same compass width, place the compass on point K and draw an arc in the interior of ∠HFK to intersect the previously drawn arc. Label the point of intersection as point J.
6. Use the straightedge to draw JH¯¯¯¯¯ and JK¯¯¯¯¯.
