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A varies directly as b as the square of c and 4-?

Mathematics

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3 parts
a
15
b
270
c

5 or −5\sqrt{5} \text{ or } -\sqrt{5}

Step-by-step solution

7 steps

Here's the solution to your variation problem.

The problem states that aa varies directly as bb and as the square of cc. This can be written as an equation:

a=kbc2a = kbc^2

where kk is the constant of variation.

We are given that a=30a=30 when b=0.5b=0.5 and c=2c=2.

Part a· 2 steps

Find the value of kk, the constant of variation.

  1. Substitute the given values into the variation equation.

    30=k(0.5)(2)230 = k(0.5)(2)^2
  2. Simplify and solve for kk.

    30=k(0.5)(4)30 = k(0.5)(4) 30=k(2)30 = k(2) k=302k = \frac{30}{2} k=15k = 15

    The constant of variation is 15\boxed{15}.

Part b270

Find the value of aa when b=2b=2 and c=3c=3.

  1. Use the constant of variation k=15k=15 found in part (a) and the new values for bb and cc.

    a=15(2)(3)2a = 15(2)(3)^2
  2. Calculate the value of aa.

    a=15(2)(9)a = 15(2)(9) a=30(9)a = 30(9) a=270a = 270

    The value of aa is 270\boxed{270}.

Part c

5 or −5\sqrt{5} \text{ or } -\sqrt{5}

Find the values of cc when a=300a=300 and b=4b=4.

  1. Use the constant of variation k=15k=15 and the given values for aa and bb.

    300=15(4)c2300 = 15(4)c^2
  2. Simplify and solve for c2c^2.

    300=60c2300 = 60c^2 c2=30060c^2 = \frac{300}{60} c2=5c^2 = 5
  3. Take the square root of both sides to find cc.

    Remember to include both positive and negative roots.

    c=±5c = \pm\sqrt{5}

    The values of cc are 5 or −5\boxed{\sqrt{5} \text{ or } -\sqrt{5}}.

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