HELP

Write the ratio in simplest form. <br> 12 DVDS to 10 books.

e-lub [12.9K]

Answer:

6:5

Step-by-step explanation:

divide both numbers by using gcd/ greatest common divisor

= 12:10/ 2

=6:5

please mark me as brainliest!!! Good luck with your tests!

8 0

1 year ago

What is the reciprocal of 1 and 2/3?

mr_godi [17]

Answer: 3/5

Step-by-step explanation: To find the reciprocal of 1 and 2/3, we first need to convert 1 and 2/3 to an improper fraction.

To write a mixed number as an improper fraction, first multiply the denominator times the whole number, then add the numerator.

So here, we multiply the denominator, 3, times the whole number, 1, then we add the numerator. So we have (3 x 1) + 2 or 3 + 2 which is 5.

Now we put our new number over our old denominator and we get 5/3.

However, we're not done yet. We still need to find the reciprocal.

The reciprocal describes a fraction so when we're asked to find the reciprocal of 5/3, we want to switch the numerator and denominator.

So if we switch 5/3, we get the proper fraction 3/5.

Therefore, the mixed number 1 and 2/3 is equivalent to the improper fraction 5/3, and if we switch the top and bottom numbers, we get 3/5.

7 0

2 years ago

PLZ HELP ASAP WILL GET BRAINLIEST WHEN IT LETS ME

solniwko [45]

That part is the median, which is 57.

7 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

1 year ago

Different ways to show subtraction for a problem like 32 - 15

chubhunter [2.5K]

15-32= 32-15= 32 -15 15 -32

8 0

2 years ago