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3 years ago

Describe how to transform the graph of g(x)= ln x into the graph of f(x)= ln (3-x) -2.

Mathematics

2 answers:

Strike441 [17]3 years ago

7 0

Answer:

The graph of g(x) = ㏑x translated 3 units to the right and then reflected

about the y-axis and then translated 2 units down to form the graph of

f(x) = ㏑(3 - x) - 2

Step-by-step explanation:

* Lets talk about the transformation

- If the function f(x) reflected across the x-axis, then the new

function g(x) = - f(x)

- If the function f(x) reflected across the y-axis, then the new

function g(x) = f(-x)

- If the function f(x) translated horizontally to the right

by h units, then the new function g(x) = f(x - h)

- If the function f(x) translated horizontally to the left

by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up

by k units, then the new function g(x) = f(x) + k

- If the function f(x) translated vertically down

by k units, then the new function g(x) = f(x) – k

* lets solve the problem

∵ Graph of g(x) = ㏑x is transformed into graph of f(x) = ㏑(3 - x) - 2

- ㏑x becomes ㏑(3 - x)

∵ ㏑(3 - x) = ㏑(-x + 3)

- Take (-) as a common factor

∴ ㏑(-x + 3) = ㏑[-(x - 3)]

∵ x changed to x - 3

∴ The function g(x) translated 3 units to the right

∵ There is (-) out the bracket (x - 3) that means we change the sign

of x then we will reflect the function about the y-axis

∴ g(x) translated 3 units to the right and then reflected about the

y-axis

∵ g(x) changed to f(x) = ㏑(3 - x) - 2

∵ We subtract 2 from g(x) after horizontal translation and reflection

about y-axis

∴ We translate g(x) 2 units down

∴ g(x) translated 3 units to the right and then reflected about the

y-axis and then translated 2 units down

* The graph of g(x) = ㏑x translated 3 units to the right and then

reflected about the y-axis and then translated 2 units down to

form the graph of f(x) = ㏑(3 - x) - 2

LekaFEV [45]3 years ago

5 0

Answer:

c edge

Step-by-step explanation:

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3 years ago

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romanna [79]

Answer:

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Step-by-step explanation:

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3 years ago

Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A balloon, rising vert

Liula [17]

Answer:

(a) 2.79 seconds after its release the bag will strike the ground.

(b) At a velocity of 73.28 ft/second it will hit the ground.

Step-by-step explanation:

We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.

Assume the acceleration of the object is a(t) = −32 feet per second.

(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;

$s=ut+\frac{1}{2} at^{2}$

Here, s = distance of the balloon above the ground = - 80 feet

u = intital velocity = 16 feet per second

a = acceleration of the object = -32 feet per second

t = required time

So, $s=ut+\frac{1}{2} at^{2}$

$-80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})$

$-80=16t-16 t^{2}$

$16 t^{2} -16t -80 =0$

$t^{2} -t -5 =0$

Now, we will use the quadratic D formula for finding the value of t, i.e;

$t = \frac{-b\pm \sqrt{D } }{2a}$

Here, a = 1, b = -1, and c = -5

Also, D = $b^{2} -4ac$ = $(-1)^{2} -(4 \times 1 \times -5)$ = 21

So, $t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}$

$t = \frac{1\pm \sqrt{21 } }{2}$

We will neglect the negative value of t as time can't be negative, so;

$t = \frac{1+ \sqrt{21 } }{2}$ = 2.79 ≈ 3 seconds.

Hence, after 3 seconds of its release, the bag will strike the ground.

(b) For finding the velocity at which it hit the ground, we will use the formula;

$v=u+at$

Here, v = final velocity

So, $v=16+(-32 \times 2.79)$

v = 16 - 89.28 = -73.28 feet per second.

Hence, the bag will hit the ground at a velocity of -73.28 ft/second.

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