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tangare [24]
3 years ago
15

Select the correct answer.

Mathematics
2 answers:
andrew-mc [135]3 years ago
5 0

Answer:

C. \displaystyle (-2, 32)\:and\:(2, -32)

Step-by-step explanation:

The best way to put this is that the integer signs MUST be interchanged.

I am joyous to assist you anytime.

vekshin13 years ago
3 0

Answer:

C

Step-by-step explanation:

Googling this, we find that when -f(x) = f(-x), our function is odd. This means that when we multiply our input by -1, our output will multiply by -1 as well.

A is incorrect because when we multiply our input (-1) by -1, our output does not change

B is incorrect because it isn't even a function -- there are multiple outputs for the input

C is correct because when we multiply the input by -1 (as -2 goes to 2), the output multiplies by -1 (32 goes to -32)

D is incorrect because when we multiply the input by -1, the output stays the same. This is an example of an even function.

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Answer:

  • Question 1a. i) x=1.8m

  • Question 1a. ii)   Volume=27.6m^3

  • Question 1b) Volume=65.9m^3

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

         tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}

For the smalll triangle you can write:

        tan(\theta )=\dfrac{x}{1m}

For tthe big triangle:

      tan(\theta )=\dfrac{x+2.7m}{2.5m}

Substitute:

        \dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}

Solve for x:

        2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

  • Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

         Volume=(1/3)\pi \times radius^2\times height

Substitute:

         Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3

  • Find the volume of a cone with heigth = 1.8m and radius = 1m

        Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3

  • Subtract the volume of the small cone from the volume of the big cone

        Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

            \dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}

           2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m

        Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)

       Volume=38.3m^3

<u>iii) Add the volume of the two frustrums</u>

  • Volume=27.6m^3+38.3m^3=65.9m^3

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Write the coefficients of the x-terms as the numbers down the first column

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