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2 years ago

HELP ME PLS I NEED THIS

Mathematics

2 answers:

Veronika [31]2 years ago

7 0

Hopefully this helps as a hint

Fittoniya [83]2 years ago

5 0

Hello me too i need this

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From first principle find the derivative of logx

Archy [21]

Hello!

$\large\boxed{\frac{dy}{dx}logx = \frac{1}{xln10} }$

Recall that:

$logx$ can be rewritten as:

$log_{10}x$

Use the equation for the derivative of a log expression:

$\frac{dy}{dx}log_{a}u = \frac{1}{ulna} * u'$

Substitute in the values in the expression:

$logx = \frac{1}{xln10} * 1 = \frac{1}{xln10}$

4 0

2 years ago

What is 6 3/4 x 3/5<br><br> A) 54/20<br><br> B) 75/12<br><br> C) 81/20<br><br> D) 9/4

steposvetlana [31]

The answer you are looking for is C my friend

8 0

3 years ago

The distance a bike travels varies directly as the amount of time the bike has been ridden. Angela is riding her bike. If she tr

Levart [38]

Answer:

420 miles

Step-by-step explanation:

60 ×7=420

thank

4 0

3 years ago

Read 2 more answers

HELP ILL GIVE BRAINLIEST

SVETLANKA909090 [29]

Answer:

$\huge\boxed{\bf\:c = \frac{11}{4}}$

Step-by-step explanation:

$4c + 2c = 4c + \frac{11}{2}$

Cancel 4c on both the sides of the equation.

$2c = \frac{11}{2}$

Bring the 2 in the denominator of 11/2 to the left hand side of the equation.

$2*2c = 11\\4c = 11$

Bring 4 to the right hand side if the equation.

$\boxed{\bf\:c = \frac{11}{4}}$

$\rule{150pt}{2pt}$

4 0

1 year 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