<u>solution</u>
W=f×d×cos theta
=25N×16m×cos 45°
=400Nm×cos 45°
=282.84J
W=f×d×cos theta
=25N×7m×cos 10°
=175Nm×cos10°
=172.34J
Total workdone=282.84J+172.34J=455.18J
Answer:
Step-by-step explanation:
Answer:
The correct option is;
21 ft
Step-by-step explanation:
The equation of the parabolic arc is as follows;
y = a(x - h)² + k
Where the height is 25 ft and the span is 40 ft, the coordinates of the vertex (h, k) is then (20, 25)
We therefore have;
y = a(x - 20)² + 25
Whereby the parabola starts from the origin (0, 0), we have;
0 = a(0 - 20)² + 25
0 = 20²a + 25 → 0 = 400·a + 25
∴a = -25/400 = -1/16
The equation of the parabola is therefore;
To find the height 8 ft from the center, where the center is at x = 20 we have 8 ft from center = x = 20 - 8 = 12 or x = 20 + 8 = 28
Therefore, plugging the value of x = 12 or 28 in the equation for the parabola gives;
.
Answer:
length: 16 m; width: 13 m
Step-by-step explanation:
Write each of the statements as an equation. You know that the formula for the perimeter is ...
P = 2(L +W)
so one of your equations is this one with the value of P filled in:
• 2L + 2W = 58
The other equation expresses the relation between L and W:
• L = W +3 . . . . . . . . the length is 3 meters greater than the width
There are many ways to solve such a system of equations. Since you have an expression for L, it is convenient to substitute that into the first equation to get ...
2(W+3) +2W = 58
4W +6 = 58 . . . . . . . simplify
4W = 52 . . . . . . . . . . subtract 6
W = 13 . . . . . . . . . . . .divide by 4
We can use the expression for L to find its value:
L = 13 +3 = 16
The length is 16 meters; the width is 13 meters.
Answer:
Point R
Step-by-step explanation:
Its the farthest away from the majority
