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AveGali [126]
2 years ago
14

6. The variable x and y vary directly, and y = 40 when x = 5. Which equation relates the variables?

Mathematics
1 answer:
valkas [14]2 years ago
5 0

Answer:

The equation relates the variables is y = 8x ⇒ D

Step-by-step explanation:

If x and y vary directly (y ∝ x), then y = k x, where

  • k is the constant of variation
  • k can be found using the initial value of x and y

∵ The variable x and y vary directly

∴ y ∝ x

→ By using the rule above

∴ y = k x

∵ y = 40 when x = 5

∴ The initial values of x and y are x = 5 and y = 40

→ Substitute them in the equation above to find k

∵ 40 = k(5)

∴ 40 = 5k

→ Divide both sides by 5

∵ \frac{40}{5} = \frac{5k}{5}

∴ 8 = k

→ Substitute the value of k in the equation above

∴ y = 8x

∴ The equation relates the variables is y = 8x

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3 0
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Liula [17]

Answer:

\boxed{\boxed{x=\dfrac{\pi}{3}\ \vee\ x=\pi\ \vee\ x=\dfrac{5\pi}{3}}}

Step-by-step explanation:

\cos(3x)=-1\iff3x=\pi+2k\pi\qquad k\in\mathbb{Z}\\\\\text{divide both sides by 3}\\\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\\\\x\in[0,\ 2\pi)

\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)

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The answer would be A. There is a phase shift to the left. Because you are adding pi/6.
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