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3 years ago

13

What is a ordered pair?

2 answers:

Alchen [17]3 years ago

6 0

Yes yes it is D: X and Y values written in the form (x,y)

olganol [36]3 years ago

5 0

D: x and y values written in the form (x,y)

You might be interested in

Find the least common denominator of 1/6 and 2/7

harina [27]

Answer:

it is it is 42

Step-by-step explanation:

7/42+12/42=19/42

5 0

3 years ago

Work out the area of a parallelogram with a base of 12cm and a height of 4cm?

olga nikolaevna [1]

It is 48 cm together because mutiply it

6 0

3 years ago

Read 2 more answers

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

Help please <br> 1 3/5 x 2 1/3=

Crank

Answer:

= 56/ 15 and 3 11/15

Step-by-step explanation:

1 3/5 x 2 1/3

= 8/5 x 7/3

3 0

3 years ago

How do I solve this?

andreev551 [17]

Answer:

x is in the range [-1,4]

Step-by-step explanation:

I haven't worked with absolute value inequalities in awhile, but let's take a wack at this.

We are given the following inequality:

| 2x - 3 | <= 5

This implies two possible cases:

[1] -5 <= 2x -3

Or

[2] 2x - 3 <= 5

So let's solve x for both of these cases:

[1] -5 <= 2x - 3

-2 <= 2x

-1 <= x

[2] 2x - 3 <= 5

2x <= 8

x <= 4

So from these cases, we can say the following is true:

x >= -1 and x <= 4

Thus, we can write this in the form

-1 <= x <= 4

Or in interval notation:

{ x is element of reals | -1 <= x <= 4}

Also written as

x is in the range [-1,4]

Where the closed brackets represent 1 and 4 as possible answers whereas parenthesis would imply they were not.

Cheers.

4 0

3 years ago