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3 years ago

Two values that have a sum of 13. Probability

Mathematics

2 answers:

Crazy boy [7]3 years ago

5 0

Answer:10 and 3

Step-by-step explanation:10 plus 3 equal 13 and those numbers are both values

GrogVix [38]3 years ago

3 0

Answer:

11 and 2

Step-by-step explanation:

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Bro whats is 3,972 divde 12

leonid [27]

Answer:

The answer would be 331.

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3 years ago

If the diameter of a microscope's field is 0.12mm, how wide is an object that fills one-quarter of the field? (Explain how you'v

vlada-n [284]

Answer:.3mm

Step-by-step explanation:

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3 years ago

If x is divided by 5 the remainder is 4. If why is divided by 5 the remainder is 1. What is the remainder when x+y is divided by

Naddik [55]

Answer:

Step-by-step explanation:

We are given the following:

$\frac{x}{5}=q_1+\frac{4}{5}$ (equ. 1)

$\frac{y}{5}=q_2+\frac{1}{5}$ (equ. 2)

We are asked:

$\frac{x+y}{5}=q_3+\frac{r}{5}$ (equ. 3) , what is $r$ ?

The $q_i$ 's represent the quotients you get.

$r$ is the remainder of dividing $x+y$ by 5.

We know that $r$ is a number in {0,1,2,3,4}.

$x=5q_1+4$ (I got this by multiplying both sides of equ 1. by 5.)

$y=5q_2+1$ (I got this by multiplying both sides of equ 2. by 5.)

Let's add these equations together:

$x+y=(5q_1+5q_2)+(4+1)$

Factoring the 5 out for the $q_i$ 's part and simplify 4+1 gives:

$x+y=5(q_1+q_2)+5$

So $5$ can't be the remainder of dividing something by 5 but see that we can factor this right hand expression more as:

$x+y=5(q_1+q_2+1)$

So $q_3=q_1+q_2+1$ while there is no remainder (the remainder is 0).

Let's do an example if you are not convinced at this point that the remainder will be 0.

So choose x=9 since 9/5 gives a remainder of 4.

And choose y=16 since 16/5 gives a remainder of 1.

x+y=9+16=25 and 25/5 gives a remainder of 0.

8 0

3 years ago

0 find the smallest values of n which will guarantee a theoretical error less than if the integral is estimated by: (a) the trap

goldenfox [79]

Answer:

Step-by-step explanation:

Check attachment for answers.

3 0

3 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

3 years ago