The legs of an isosceles right triangle increase in length at a rate of 2 m/s. determine each. (a) at what rate is the area of the triangle changing when the legs are 2 m long? (b) at what rate is the area of the triangle changing when the hypotenuse is 1 m long? (c) at what rate is the length of the hypotenuse changing?

a) The area of the triangle changing when the legs are 2 m long is 4 meter square.

b) The area of the triangle changing when the hypotenuse is 1 m long is meter square.

c) The rate is the length of the hypotenuse changing is m/s.

Step-by-step explanation:

Given : The legs of an isosceles right triangle increase in length at a rate of 2 m/s.

Let the length of a leg of an isosceles right triangle be 'x'.

Let the length of the hypotenuse be 'h'.

The are of the triangle is given by,

....(1)

Differentiate w.r.t t,

We have given,

(differentiate (1) w.r.t x)

....(2)

a) The legs are 2 m long  i.e. x=2 m

Substitute in (1),

The area of the triangle changing when the legs are 2 m long is 4 meter square.

b) The hypotenuse is 1 m long.

Applying Pythagoras theorem,

Put h=1,

Substitute the value of x in equation (2),

The area of the triangle changing when the hypotenuse is 1 m long is meter square.

c) We know,

or

Derivate w.r.t t,

We have given,

So,

The rate is the length of the hypotenuse changing is m/s.

x=6 and x=3

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g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.