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Dmitrij [34]
3 years ago
15

2.

Mathematics
1 answer:
victus00 [196]3 years ago
7 0

Answer:

With a vertex (h, k) at (0, 3) and given that  a = -3, then the equation of this parabola in vertex form is as follows:

    y = a(x - h)2 + k

    y = -3(x - 0)2 + 3

    y = -3x2 + 3

Step-by-step explanation:

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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).
Aleksandr-060686 [28]

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}

5 0
3 years ago
Triangle ABC is such that AB = 3cm , BC = 4cm , angle ABC = 120 and BAC = theta : Write down in terms of theta , and expression
lakkis [162]
1)inside the triangule is 180º .There are 3 angules, one with 120º, theta and
angle ACB, lets call it x.

x + theta + 120 = 180
x + theta= 180-120
x + theta= 60
x= 60- theta
angle ACB= 60- theta
8 0
3 years ago
Explain how 7/12 is greater than 1/3 but less than 2/3
Zepler [3.9K]
First we will change them on the same denominator which will be 12. If we do something to the denominator we must do the same to the numerator so :
For 1/3 we get  4/12 because (1/3)*4 = 4/12
And for 2/3 we get 8/12 because (2/3)*4 = 8/12
So 1/3 is the smaller fraction, 7/12 is in the middle and 2/3 is the bigger fraction.
6 0
3 years ago
Read 2 more answers
Please help
elena-14-01-66 [18.8K]

Brainly is so proper why is no one saying inappropriate things?

8 0
2 years ago
Read 2 more answers
If f(x)=2x²+5√(x-2) , complete the following statement: f(3)=___
aliya0001 [1]
F(x)=2x²+5√(x-2)

f(3)=2(3)²+5√(3-2)
= 2 (9) + 5√1
= 18 + 5
= 23.
4 0
3 years ago
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