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timofeeve [1]
3 years ago
9

ASAP! find the slope of a line that is perpendicular to y=-1/2x+4

Mathematics
1 answer:
aliya0001 [1]3 years ago
8 0

We know that :

Ф  Product of the slopes of two lines perpendicular to each other should be equal to -1

Let the slope of the line perpendicular to given line be : M

Given equation of the line : y = -1/2x + 4

This line is in the form : y = mx + c, where m is the slope and c is the y intercept

Comparing with y = mx + c :

we can notice that slope of the given line is -1/2

⇒   M × -1/2 = -1

⇒   M × 1/2 = 1

⇒   M = 2

<u>Answer</u> : Slope of the line perpendicular to given line is 2

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A cube has a volume of 64. What is the length of one side of the cube?
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Step-by-step explanation:

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  • The cube root of a number is the number you multiply by itself 3 times to get another number.
  • One common cube is 27, and the cube root of it is 3 because 3 x 3 x 3 = 27.

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We found the cube root of 64, or the side of the cube with a volume of 64. The side of the cube, therefore, is 4.

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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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