It would be nonsense, as if the beetle were at normal size, then it would be considered 1/1 as a scale. If you wanted to increase it, you would increase the numerator.
Answer:
t = 2.52 seconds
Step-by-step explanation:
h=139-15t-16t^2
We want to know when the ball hits the ground
That would be when h=0
0 = 139-15t-16t^2
We can use the quadratic formula to find t
t = -b ± sqrt(b^2-4ac)
----------------------
2a
where a = -16 b = -15 and c = 139
t = -(-15) ± sqrt((-15)^2-4(-16)139)
----------------------
2(-16)
t = (15) ± sqrt(225+8896)
----------------------
-32
t = (15) ± sqrt(9121)
----------------------
-32
t = 15+ sqrt(9121) t = 15- sqrt(9121)
-------------------- or -------------------
-32 -32
-3.453247707 or 2.515747707
Since time cannot be negative
2.515747707
Round to the nearest hundredth
t = 2.52 seconds
<span>(a) 5(3m+2) = 15m + 10
(b) 6(g+h) = 6g + 6h
(c) 4d+8 = 4(d + 2)
(d) 21p+35q = 7(3p + 5q)
(e) 18x+9y = 9(2x + y)</span>
Answer:
r = 5
Step-by-step explanation:
To solve for r, plug in the change in x values and change in y values, since the hypotenuse is simply the diagonal of both the y-coordinate and x-coordinates shown via drawing legs on the vertical and horizontal axis. So, since (0,0) is the initial point, r = sqr[(-4-0)^2 + (-3-0)^2 = sqr(16 + 9) = 5.
Now, the angle theta is the angle in which the sine, cosine, and tangent ratios are found. Simply use opposite over hypotenuse for sine, adjacent over hypotenuse for cosine, and opposite over adjacent for tangent using theta as the angle in which these values are obtained.
The answer to the question is
20/30. 24/30,. 21/30