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3 years ago

14

Find the value of given expression

itle=" \sqrt{440 + 1} " alt=" \sqrt{440 + 1} " align="absmiddle" class="latex-formula">

2 answers:

fomenos3 years ago

6 0

Answer:

21

Step-by-step explanation:

$\sqrt{440 + 1} \\ = \sqrt{441 } \\ = 21$

IceJOKER [234]3 years ago

5 0

Answer:

your answer is 21

Step-by-step explanation:

I¨m smart man XD

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Write a phrase in words to match each expression.<br> 5+3<br> —— <br> n

MA_775_DIABLO [31]

Answer:

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Step-by-step explanation:

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3 years ago

Explain how to solve 4^x+3 = 7 using the change of base formula log base b of y equals log y over log b. Include the solution fo

Masja [62]

The value of x is 1

Explanation:

The equation is $4^x+3=7$

Subtracting both sides by 3, we get,

$4^x=4$

Taking log on both sides, we get,

$\log 4^{x}=\log 4$

Rewriting the equation by $4^{x}=u$

Thus, we have,

$\log u=\log 4$

Applying log rule, if $\log f(x)=\log g(x)$ , then $f(x)=g(x)$

Thus, $u=4$

Substituting $u=4$ in $4^{x}=u$ , we get,

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Also, since, $a^{f(x)}=a^{g(x)}$ , then $f(x)=g(x)$

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7 0

3 years ago

Find y' if x = tan(y)

Travka [436]

Use the chain rule:

$x=\tan y$
$\dfrac{\mathrm d}{\mathrm dx}x=\dfrac{\mathrm d}{\mathrm dx}\tan y$
$1=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$
$1=\sec^2y\,y'$
$y'=\dfrac1{\sec^2y}=\cos^2y$

Furthermore, if you restrict the domain of $y$ , we could write

$x=\tan y\iff \tan^{-1}x=y$

and so we could also have

$y'=\cos^2(\tan^{-1}x)=\left(\dfrac1{\sqrt{x^2+1}}\right)^2=\dfrac1{x^2+1}$

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4 years ago

A car drives at a speed of 90 km/h for two hours and 20 minutes how far does the car drive?

lesya [120]

Answer:

210km

Step-by-step explanation:

1h...........90 km

2h...........2*90=180 km

20min=60min/3.........90km/3=30 km

2h 20min.......180+30=210km

3 0

3 years ago

Read 2 more answers