The length of a rectangle is increasing at a rate of and its width is increasing at a rate of . when the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?

220 cm2/s 
 EXPLINATION

L length of rectangel (cm)

W width of rectangel(cm)

A area of rectangel (cm²)

T time (s)

we are told that:

d1\dt=8cm\s(const),and dw\dt =3cm\s(const)

the area of the rectangel is

A=LW

Differentiating wrt t (using the product rule) we get;

dAdt=(l)(dwdt)+(dldt)(w)

∴dAdt=3l+8w

So when l=20 and w=10⇒
     dAdt=3⋅20+8⋅20
∴dAdt=60+160
∴dAdt=220 cm2/s

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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