Answer:
The period of the sine curve is the length of one cycle of the curve. The natural period of the sine curve is 2π. So, a coefficient of b=1 is equivalent to a period of 2π. To get the period of the sine curve for any coefficient b, just divide 2π by the coefficient b to get the new period of the curve.
Step-by-step explanation:
Answer:
Average rate of change = 20.2
Rate of change at the left endpoint : f' (5) = 4t = 20
Rate of change at the right endpoint : f' (5.1) = 4*5.1 = 20.4
Step-by-step explanation:
The average rate of change of the function
F(t) = 2t^2 - 1 , [ 5,5.1 ]
solution
= 
= [ 2 ( 26.01 - 25 ) / 0.1 ]
= 2.02 / 0.1 = 20.2
Rate of change at the left endpoint : f' (5) = 4t = 20
Rate of change at the right endpoint : f' (5.1) = 4*5.1 = 20.4
DEF= 6 ft^2
A= 1/2*b*h
Plug in what you know.
6= 1/2*4*h
Multiply like terms.
6=2*h
Divide 2 on both sides.
3=h
If the base of ABC= 12
12/4=3
3*3=9
A= 1/2*b*h
A= 1/2*12*9
A= 1/2*84
A= 42
The area of ABC is 42 ft^2
I hope this helps!
~kaikers
Answer:
L2: y-0 = 5/2(x-5)
y = 5/2x-25/2
Step-by-step explanation:
Parallel lines have same slopes.
Line 1, L1: 5x-2y=20 is in standard form Ax+By=C therefore slope m1= -A/B = -5/-2 = 5/2 or you can solve it for y so you will have the equation in slope-intercept form.
5x-2y = 20
-2y = -5x+20
y = (-5/-2)x+20/(-2)
y = (5/2)x-10 hence m1=5/2 and y-intercept is -10
Line 2 , L2: y-y1 = m (x-x1), m=m2=m1=5/2
Point p(5,0) or p(x1,y1) therefore x1=5 , y1=0 and m=5/2
L2: y-0 = 5/2(x-5)
y = 5/2x-25/2
