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joja [24]
4 years ago
4

Which function results after applying the sequence of transformations to f(x)=x^5

Mathematics
2 answers:
Murljashka [212]4 years ago
8 0

Answer:

A is very correct

Step-by-step explanation:

Gemiola [76]4 years ago
4 0

Answer:

The final function is g(x)=\dfrac{1}{2}(x+2)^5-1

A is correct

Step-by-step explanation:

Given: Parent function f(x)=x^5

We need to apply sequence of transformation.

Step 1: Compress vertically by 1/2

If function compress vertically number multiply by factor

f(x)=\dfrac{1}{2}x^5

Step 2: Shift 2 unit left

For left and right shift change in horizontal.

For a unit change , x-> x+a

f(x)=\dfrac{1}{2}(x+2)^5

Step 3: Shift 1 unit down

For up and down change in y value or vertically shift.

For down subtract 1 unit from function

f(x)=\dfrac{1}{2}(x+2)^5-1

Please see the attachment for transformation step by step.

Hence, The final function is f(x)=\dfrac{1}{2}(x+2)^5-1

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Neporo4naja [7]

a^2 + b^2 = c^2

a = sqrt(c^2 - b^2) = sqrt (20^2 - 16^2)

= 12

sin\theta = a/c = 12/20

= 3/5

cos\theta = b/c = 16/20

= 4/5

tan\theta = a/b = 12/16

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sec\pi = c/b = 20/16

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= 1 1/4


4 0
3 years ago
Read 2 more answers
3. Jessica is awake 2/3 of the day. She spends 5/8 of this time at home
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Answer:  a. 5/12     b. 10 hours

Step-by-step explanation:

5/8 x 2/3 = 10/24 = 5/12

she spends 5/12 of the day awake at home

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so she spends 10 hours awake at home

7 0
4 years ago
g Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min, and its coarseness is such that it forms a pile in the sh
Maslowich

Answer:

0.3537 feet per minute.

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

\dfrac{dV}{dt}=10$ ft^3/min

\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h

If the Base Diameter = Height of the Cone

The radius of the Cone = h/2

Therefore,

\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}

\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}

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When h=6$ feet$\\\dfrac{3\pi *6^2}{12}\dfrac{dh}{dt}=10\\9\pi \dfrac{dh}{dt}=10\\ \dfrac{dh}{dt}= \dfrac{10}{9\pi}\\ \dfrac{dh}{dt}=0.3537$ feet per minute

When the pile is 6 feet high, the height of the pile is increasing at a rate of 0.3537 feet per minute.

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