All categories

3 years ago

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:

You might be interested in

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

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

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

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:

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

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