Answer:
Step-by-step explanation:
hello :
The circumference of the circle is 6π inches
The length of the arc is (340/360)×6π = 17.8 inches
(The complete circumference would cover 360°. Angle ALB is 340°)
:
<span>Constraints (in slope-intercept form)
x≥0,
y≥0,
y≤1/3x+3,
y</span>≤ 5 - x
The vertices are the points of intersection between the constraints, or the outer bounds of the area that agrees with the constraints.
We know that x≥0 and y≥0, so there is one vertex at (0,0)
We find the other vertex on the y-axis, plug in 0 for x in the function:
y <span>≤ 1/3x+3
y </span><span>≤1/3(0)+3
y = 3.
There is another vertex at (0,3)
Find where the 2 inequalities intersect by setting them equal to each other
(1/3x+3) = 5-x Simplify Simplify Simplify
x = 3/2
Plugging in 3/2 into y = 5-x: 10/2 - 3/2 = 7/2
y=7/2
There is another vertex at (3/2, 7/2)
There is a final vertex where the line y=5-x crosses the x axis:
0 = 5 -x , x = 5
The final vertex is at point (5, 0)
Therefore, the vertices are:
(0,0), (0,3), (3/2, 7/2), (5, 0)
We want to maximize C = 6x - 4y.
Of all the vertices, we want the one with the largest x and smallest y. We might have to plug in a few to see which gives the greatest C value, but in this case, it's not necessary.
The point (5,0) has the largest x value of all vertices and lowest y value.
Maximum of the function:
C = 6(5) - 4(0)
C = 30</span>
B is the correct answer choice,you didn’t hear it from me
From the beginning we know we are dealing with two different rates. Distance is equal to (Speed X Time). We have two different speeds though. We will call the min speed = x and the max speed = y. With those two ideas in place we can make two equations for both Tony and Rae.
Tony:
2x+3.5y=355
Rae:
2x+3y=320
Now you can solve the equations by subtracting Rae's equation from Tony's
.5y=35 solve for y and you get 70
Now plug that y value back into either equation and you can solve for x. I'll use Rae's distance:
2x+3(70)=320
x=55
Answer:
So you would not be carried away by the tides
Step-by-step explanation:
hope this is what u want
