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pickupchik [31]
3 years ago
6

A rectangle has a length that is increasing at a rate of 10 mm per second with the width being held constant. What is the rate o

f change of the area of the rectangle if the width is 8 mm?
Mathematics
1 answer:
NISA [10]3 years ago
6 0

Answer:

the rate of change of the area is Ra= 80 mm² per second

Step-by-step explanation:

the area of a rectangle (A) is

A = L * W , L= length and W= width

if the width remains constant the change in the area is only due to the change in length , thus:

ΔA = Δ( L * W ) = W * ΔL , where ΔA represents the change in area and ΔL represents the change in length

ΔA = W ΔL

denoting Δt as the time required to change, and dividing both sides by Δt

ΔA/Δt = W ΔL/Δt

where Ra=ΔA/Δt = rate of change of the area and RL=ΔL/Δt rate of change of the length.

thus

Ra = W * RL

replacing values

Ra = W * RL = 8 mm * 10 mm/second = 80 mm² /second

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Answer:

\huge\boxed{Center = (3,3),Radius = 2\sqrt{15} }

Step-by-step explanation:

<u><em>Given equation is</em></u>

x^2 + y^2 -6x-6y -42 = 0

Adding 42 to both sides

x^2 + y^2 -6x-6y = 42\\

Completing squares

x^2 -6x+y^2 -6x = 42\\(x)^2 - 2(x)(3) +y^2 - 2(y)(3) = 42

Adding (3)² => 9 and (3)² => 9 to both sides

(x-3)^2+(y-3)^2 = 42+9+9\\(x-3)^2 + (y-3)^2 = 60\\(x-3)^2  (y-3)^2 = (2{\sqrt{15})^2}

Comparing it with (x-h)^2+(y-k)^2 = r^2 where Center = (h,k) and Radius = r

We get:

Center = (3,3)

Radius = 2\sqrt{15}

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answer is attached with solution

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-Bella :)
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