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Aliun [14]
2 years ago
9

Suppose that y varies directly as the square of x, and that y=567 when x=9. What is y when x=4

Mathematics
1 answer:
Gwar [14]2 years ago
6 0

Answer:

y  = 112

Step-by-step explanation:

If  y varies directly as the square of x

mathematically;

y ∝ x²

y = kx² where 'k' is the constant of proportionality

let u know the value of k for this equation by making k the subject of the formular from y = kx²

divide both sides by x²

k = y / x²

given that  y = 567 x = 9

k = 567/(9)²

k = 567/81

k = 7

now to get the value of  y when x = 4

from the equation conneting x, y,

y = kx²

y = 7 × (4)²

y = 7 × 16

y  = 112

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I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
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Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

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