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White raven [17]

3 years ago

13

Isabella has dance class every 3 days and soccer

1 answer:

Nataliya [291]3 years ago

4 0

Answer:

15

Step-by-step explanation:

first you need to find the LCM (least common multiple) which is 15 so then every 15 days she will have both soccer, dance class and swimming class on the same day

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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

SOMEONE PLS HELP!<br> (4+51)(4 - 51) =<br> DONE

Gala2k [10]

The right answer is -2585

4 0

2 years ago

Add. write your answer in simplest form.<br> -4 _/135 + 7 _/15

Zarrin [17]

$\huge\quad\qquad\underline{{\sf Answer}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: - 4 \sqrt{135} + 7 \sqrt{15}$

$\qquad \tt \dashrightarrow \: - 4 \sqrt{5 \times3 \times 3 \times 3} + 7 \sqrt{15}$

$\qquad \tt \dashrightarrow \:( - 4 \times 3) \sqrt{5 \times 3} + 7 \sqrt{15}$

$\qquad \tt \dashrightarrow \: - 12 \sqrt{15 } + 7 \sqrt{15}$

$\qquad \tt \dashrightarrow \: - 5 \sqrt{15}$

7 0

2 years ago

Help me with this problem?<br> 15 points

kobusy [5.1K]

Answer:

See explanation

Step-by-step explanation:

Since 12 is 3 times greater than 4, you simply need to divide all of the ingredients by 3.

Cactus chunks: 4 1/2 divided by 3 is 3/2 or 1 1/2

Crushed Tumbleweed: 8/3 is just 8/3 or 2 2/3

Cactus Juice: 10/4 or 2 1/2

Lizard eggs: 4

Crumbled Flower Petals: 7/4 or 1 3/4

Water: 10/3 or 3 1/3

Hope this helps!

5 0

3 years ago

Identify the vertex and

stiv31 [10]

okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk what the queston again lol I forgt

5 0

2 years ago