Maybe I'm not understanding, but I would say 32 ounces divided by 3 gives an equal amount of food for each dog. So i would just solve 32/3. Hope this helps... sorry if it didn't!
First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

(If you were to plot the actual curve, you would have both
and
, but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)
The arc length is then given by the definite integral,

We have

Then in the integral,

Substitute

This transforms the integral to

and computing it is trivial:

We can simplify this further to

Complete question :
Anand needs to hire a plumber. He's considering a plumber that charges an initial fee of $65 along with an
hourly rate of $28. The plumber only charges for a whole number of hours. Anand would like to spend no more than $250, and he wonders how many hours of work he can afford.
Let H represent the whole number of hours that the plumber works.
1) Which inequality describes this scenario?
Choose 1 answer:
A. 28 + 65H < 250
B. 28 + 65H > 250
C. 65 + 28H < 250
D. 65 +28H > 250
2) What is the largest whole number of hours that Anand can afford?
Answer:
65 + 28H < 250
Number of hours Anand can afford = 6 hours
Step-by-step explanation:
Given the following information :
Initial hourly rate = $65
Hourly rate = $28
Number of hours worked (whole number) = H
Maximum budgeted amount to spend = $250
Therefore ;
(Initial charge + total charge in hours) should not be more than $250
$65 + ($28*H) < $250
65 + 28H < 250
Number of hours Anand can afford :
65 + 28H < 250
28H < 250 - 65
28H < 185
H < (185 / 28)
H < 6.61
Sinve H is a whole number, the number of hours he can afford is 6 hours
No such thing as cot (61,4). I think you meant to say cot (61.4°). See the difference?
So, cot (61.4°) is approximately -0.139844.
Rounded to 3 decimal places we get -0.140.
Answer:
C) 0 <\ x OR x />7
Step-by-step explanation:
The first arrow has a closed circle which indicates greater/less than or equal to. All the values shaded are to the left of 0. So, <em>x</em> is less than or equal to 0. The second arrow also has a closed circle. Since the shaded part of the arrow is to the right of 7, <em>x</em> is greater than or equal to 7.