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notka56 [123]
3 years ago
6

A wheel has a radius of 9 inches and travels as it spins. How far does the wheel travel in one revolution? Use 3.14 for pi and r

ound to the nearest inch.
Mathematics
1 answer:
Naddika [18.5K]3 years ago
7 0
I think it will travel in one revolution is 56.52
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Please help! files attached.
kotykmax [81]
The first question would be x + (x + 4). This is because Max has $x and Keisha has (x + 4) because she has four MORE THAN Max. I am pretty sure that the second one is A, but I am not positive. The first one is right though. I hope this helps!!!
5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B2x%7D%20%2B%2012%20%3D%20x%20%2B%208%20" id="TexFormula1" title=" \sqrt{2x} + 12
Helga [31]
First note that x\ge0; otherwise \sqrt{2x} is undefined as long as we're only looking for real solutions.

We can write

\sqrt{2x}+12=x+8
\sqrt{2x}=x-4
2x=(x-4)^2
2x=x^2-8x+16
x^2-10x+16=0
(x-8)(x-2)=0

\implies x=8,x=2
4 0
3 years ago
You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling
BARSIC [14]

Answer:

  (a) 15 m

  (b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

  d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

  d(2) = 25 - (5/2)·4 = 15 . . . . meters

__

(b) The <em>relative velocity</em> at 2.4 seconds is ...

  d'(t) = -5t

  d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

__

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

  d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

  10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

  (110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

     = 26 5/36 m/s = 94.1 km/h

_____

<em>Comment on following distance</em>

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0
3 years ago
3 b. A taxi hurries 261 km in 135 minutes. What is its average speed in kilometers per hour?​
otez555 [7]
Speed=d/t
Ans-1.93
Correct me if I’m wrong
3 0
3 years ago
What point in the feasible region maximizes the objective function, 3x + y ≤ 12, x+y ≤5, x ≥0,y ≥0
forsale [732]

Step-by-step explanation:

We have to find the point in the feasible region which maximizes the objective function. To find that point first we need to graph the given inequalities to find the feasible region.

Steps to graph 3x + y ≤ 12:

First we graph 3x + y = 12 then shade the graph for ≤.

plug any value of x say x=0 and x=2 into 3x + y = 12 to find points.

plug x=0

3x + y = 12

3(0) + y = 12

0 + y = 12

y = 12

Hence first point is (0,12)

Similarly plugging x=2 will give y=6

Hence second point is (2,6)

Now graph both points and joint them by a straight line.

test for shading.

plug any test point which is not on the graph of line like (0,0) into original inequality 3x + y ≤ 12:

3(0) + (0) ≤ 12

0 + 0 ≤ 12

0 ≤ 12

Which is true so shading will be in the direction of test point (0,0)


We can repeat same procedure to graph other inequalities.

From graph we see that ABCD is feasible region whose corner points will result into maximum or  minimu for objective function.

Since objective function is not given in the question so i will explain the process.

To find the maximum value of objective function we plug each corner point of feasible region into objective function. Whichever point gives maximum value will be the answer

7 0
3 years ago
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