-3x-2.5=y would be an equivalent to that equation
Answer:
The intial value is 1.25
Step-by-step explanation:
Answer:
The quantity of clover seeds the worker will add is 11.14 pounds
Step-by-step explanation:
Let
x=Poppy seed
y=clover seed
24x+13y=20.70(x+y)
x=26 pounds
Substitute x=26 pounds into the equation
24x+13y=20.70(x+y)
24(26)+13y=20.70(26+y)
624+13y=538.2 +20.70y
Collect like terms
624-538.2=20.70y-13y
85.8=7.7y
Divide both sides by 7.7
y=85.8/7.7
=11.14 pounds
Therefore, the quantity of clover seeds the worker will add is 11.14 pounds
<span>Position at t=0.35s is 0.2 m
Velocity at t = 0.35s is -0.2 m/s
Since this is college level mathematics, the use of the word "acceleration" should indicate to you that you've been given the 2nd derivative of a function specifying the location of point a. And since you've been asked for the velocity, you know that you want the 1st derivative of the function. And since you've also been asked for the position, you also want the function itself. So let's calculate the desired anti-derivatives.
f''(t) = -1.08 sin(kt) - 1.44 cos(kt)
The integral of f''(t) with respect to t is:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C
In order to find out what C is, we know that at time t=0, v = 0.36 m/s. So let's plug in the values and see what C is:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C
0.36 = (1.08 cos(3*0) - 1.44 sin(3*0))/3 + C
0.36 = (1.08 cos(0) - 1.44 sin(0))/3 + C
0.36 = (1.08*1 - 1.44*0)/3 + C
0.36 = 0.36 + C
0 = C
So the first derivative will be f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k
Now to get the actual function by integrating again. Giving:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C
And let's determine what C is:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C
0.16 = (1.08 sin(3*0) + 1.44 cos(3*0))/3^2 + C
0.16 = (1.08 sin(0) + 1.44 cos(0))/9 + C
0.16 = (1.08*0 + 1.44*1)/9 + C
0.16 = 1.44/9 + C
0.16 = 0.16 + C
0 = C
So C = 0 and the position function is: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2
So now, we can use out position and velocity functions to get the desired answer:
Position:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2
f(t) = (1.08 sin(3*0.35) + 1.44 cos(3*0.35))/3^2
f(t) = (1.08 sin(1.05) + 1.44 cos(1.05))/9
f(t) = (1.08*0.867423226 + 1.44*0.497571048)/9
f(t) = (0.936817084 + 0.716502309)/9
f(t) = 1.653319393/9
f(t) = 0.183702155
So the position of point a at t=0.35s is 0.2 m
Now for the velocity:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k
f'(t) = (1.08 cos(3*0.35) - 1.44 sin(3*0.35))/3
f'(t) = (1.08 cos(1.05) - 1.44 sin(1.05))/3
f'(t) = (1.08*0.497571048 - 1.44*0.867423226)/3
f'(t) = (0.537376732 - 1.249089445)/3
f'(t) = -0.711712713/3
f'(t) = -0.237237571
So the velocity at t = 0.35s is -0.2 m/s</span>
Answer:
a reflection of ΔRST across the line y = –x
Step-by-step explanation:
A reflection across the line y = –x transforms point (x, y) into (-y, -x)
After reflecting ΔRST across the line y = –x we get:
R (-1, 3) -> (-3, 1)
S (3,-2) -> (2, -3)
T (1, -4) -> (4, -1)
where S is at the desired vertex
