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3 years ago

X^2 + 6x + 5 show equation

Mathematics

1 answer:

alekssr [168]3 years ago

5 0

Answer:

$\boxed{\sf \ \ \ x=-5 \ \ \ or \ \ \ x=-1 \ \ }$

Step-by-step explanation:

$x^2+6x+5 = x^2+x+5x+5 = x(x+1)+5(x+1)=(x+5)(x+1)$

$x^2+6x+5 = 0 (x+5)(x+1)=0\\\\ x = -5 \ or \ x=-1$

hope this helps

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Problem number 26 of the Rhind Papyrus says: Find a quantity such that when it is added to StartFraction 1 Over 4 EndFraction of

STALIN [3.7K]

Answer:

The quantity is 12

Step-by-step explanation:

Here, we want to find the solution to the equation.

Let the quantity we are looking for be x;

According to the question;

The quantity plus a quarter of the quantity = 15

x + 1/4(x) = 15

x + x/4 = 15

Multiply through by 4

4x + (x/4)*4 = 15 * 4

4x + x = 60

5x = 60

x = 60/5

x = 12

5 0

3 years ago

Read 2 more answers

One pipe can fill a tank in 12 hours and another can fill the tank in 8 hours. How long would it take to fill the tank using bot

JulijaS [17]

Answer:

It will take 4 hours and 48 minutes with both pipes.

Step-by-step explanation:

Add their rates of filling

1st pipe's rate = ( 1 tank ) / ( 12 hrs )

2nd pipe's rate = ( 1 tank ) / ( 8 hrs )

Let, t = hrs to fill tank using both pipes

(1 / 12) + (1 / 8) = (1 / t)

Multiply both sides by 24t

2t + 3t = 24

5t = 24

t = 4.8

0.8 x 60 = 48

It will take 4 hours and 48 minutes with both pipes

7 0

3 years ago

Please Help!!!!!!<br> I need the answer ASAP!

Evgesh-ka [11]

Bro use Socratic that will help you.

6 0

3 years ago

<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

2 years ago

Can someone help me? i really need it :)

garri49 [273]

I hope you can understand my messy handwriting ‍

3 0

2 years ago

X^2 + 6x + 5 show equation​

X^2 + 6x + 5 show equation