Answer: (b)
x g(x)
1 -16
2 -12
3 -8
If g(x) is 4*f(x), then we can find g(x) by multiplying 4 by x-5
g(x) = 4(x-5)
= 4x-20
Now we can plug in 1,2, and 3 for x to see which table makes sense.
g(1) = 4(1) - 20
= -16
g(2) = 4(2) - 20
= -12
g(3) = 4(3) - 20
= -8
hope this helps
Answer:
see explanation
Step-by-step explanation:
Under a reflection in the y-axis
a point (x, y ) → (- x, y )
Hence
A'(-5, - 3 ) → A''(5, - 3 )
B'(4, 4 ) → B''(- 4, 4 )
C'(4, 2 ) → C''(- 4, 2 )
D'(- 2, - 1 ) → D''(2, - 1 )
You are adding s^2 to your 128 ft^2. s^2 is a common representation of a square. So you could think of it as adding an area of a square of side "s" to your deck.
So one way of drawing it is to draw a rectangle with area 128 ft^2 and attaching a square to one of its side. The square that is attached has a side lenght of "s."
So, first, you have to find the smallest number that is divisible by both 12 and 10. Which is 60. So, you need 5 boxes of trophies and 6 boxes of stands
Now to get how much it will cost you now multiply the respective costs by the amount of boxes so you get
5 x 10 = 50
6 x 6 = 36
Then you add both
50 + 36 = 86
He will spend a total of $86 on both trophies and stands.
Distance of each track are:
D₁ = 428.5 yd
D₂ = 436.35 yd
D₃ = 444.20 yd
D₄ = 452.05 yd
D₅ = 459.91 yd
D₆ = 467.76 yd
D₇ = 475.61 yd
D₈ = 483.47 yd
<u>Explanation:</u>
Given:
Track is divided into 8 lanes.
The length around each track is the two lengths of the rectangle plus the two lengths of the semi-circle with varying diameters.
Thus,
Starting from the innermost edge with a diameter of 60yd.
Each lane is 10/8 = 1.25yd
So, the diameter increases by 2(1.25) = 2.5 yd each lane going outward.
So, the distances are:
D₁ = 240 + π (60) → 428.5yd
D₂ = 240 + π(60 + 2.5) → 436.35 yd
D₃ = 240 + π(60 + 5) → 444.20 yd
D₄ = 240 + π(60 + 7.5) → 452.05 yd
D₅ = 240 + π(60 + 10) → 459.91 yd
D₆ = 24 + π(60 + 12.5) → 467.76 yd
D₇ = 240 + π(60 + 15) → 475.61 yd
D₈ = 240 + π(60 + 17.5) → 483.47 yd
