Answer:
Width is 10ft and length is 6ft.
Step-by-step explanation:
The area of a shape is the amount inside a shape. It is found using multiplication or A = l*w. Since the width is w and the length is 4ft less than its width or w - 4, use these expressions in the area formula to find their amount.

Subtract 60 from both sides and factor the quadratic to solve for w.

Set each factor equal to 0 and solve.
w - 10 = 0 so w = 10
w + 6 = 0 so w = -6.
Since w represents a length or distance it cannot be negative, this means the width is 10. The length which is w - 4ft is 10 -4 = 6. The length is 6 ft.
Answer:
1 495 500 505 500
2 525 515 505 515
3 470 480 460 470
Step-by-step explanation:
Suppose to be sample are
- 495 500 505 500
- 525 515 505 515
- 470 480 460 470
In the given sample if sample 3 is difficult to control .Because engineer wants to construct a sample mean controlling the service to his company produces.In sample set first two sets have continuous four headlamps but in third it is difficult to handle the control.
the answer for this question is x=4
Answer:
Step-by-step explanation:
<u>Area of circle:</u>
<u>Circumference of circle:</u>
<u>Given</u>
<u>Find r using the second formula:</u>
- 2πr = 55
- 2*3.14r = 55
- r = 55/6.28
- r = 8.8 (rounded)
<u>Find area using the first formula:</u>
- A = 3.14*8.8²
- A = 243 m² (rounded)
The dP/dt of the adiabatic expansion is -42/11 kPa/min
<h3>How to calculate dP/dt in an adiabatic expansion?</h3>
An adiabatic process is a process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression
Given b=1.5, P=7 kPa, V=110 cm³, and dV/dt=40 cm³/min
PVᵇ = C
Taking logs of both sides gives:
ln P + b ln V = ln C
Taking partial derivatives gives:

Substitutituting the values b, P, V and dV/dt into the derivative above:
1/7 x dP/dt + 1.5/110 x 40 = 0
1/7 x dP/dt + 6/11 = 0
1/7 x dP/dt = - 6/11
dP/dt = - 6/11 x 7
dP/dt = -42/11 kPa/min
Therefore, the value of dP/dt is -42/11 kPa/min
Learn more about adiabatic expansion on:
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