Math, asked by SHIVA72552y, 2 months ago

the value of x^2x+ x(x^x) when x = 2 is​

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Answers

Answered by panchajit143
1

Step-by-step explanation:

\sf\small\underline\red{Given:-}

Given:−

\tt{\implies Radius\:_{(two\:circle)}=6cm\:and\:10cm}⟹Radius

(twocircle)

=6cmand10cm

\sf\small\underline\red{To\:Find:-}

ToFind:−

\tt{\implies The\:radius\:_{(new\:circle)}=?}⟹Theradius

(newcircle)

=?

\sf\small\underline\red{Solution:-}

Solution:−

To calculate the radius of new circle at first we have to assume the radius of 1st circle be r¹ and the radius of 2nd circle be r². The radius of new circle be r. According to given question we have to set up equation. Then applying the formula of area of circle to calculate the radius of new circle. Here the radius of 1st circle be 10cm and the radius of 2nd circle be 6cm.

\sf\small\underline\red{Formula\:used:-}

Formulaused:−

\tt{\implies Area\:of\:circle=\pi\:r^2}⟹Areaofcircle=πr

2

\tt{\implies Difference\:_{(area\:of\:2\: cicle)}=Area\:_{(new\:circle)}}⟹Difference

(areaof2cicle)

=Area

(newcircle)

\tt{\implies \pi\:(r^1)^2-\pi\:(r^2)^2=\pi\:r^2}⟹π(r

1

)

2

−π(r

2

)

2

=πr

2

\tt{\implies \pi\:(10)^2-\pi\:(6)^2=\pi\:r^2}⟹π(10)

2

−π(6)

2

=πr

2

\tt{\implies \pi(10^2-6^2)=\pi\:r^2}⟹π(10

2

−6

2

)=πr

2

\tt{\implies \pi(100-36)=\pi\:r^2}⟹π(100−36)=πr

2

\tt{\implies \pi(64)=\pi\:r^2}⟹π(64)=πr

2

\tt{\implies r^2=64}⟹r

2

=64

\tt{\implies r=\sqrt{64}}⟹r=

64

\tt{\implies r=8cm}⟹r=8cm

\sf\large{Hence'}Hence

\tt{\implies The\:radius\:_{(new\:circle)}=8cm}⟹Theradius

(newcircle)

=8cm

\sf\small\underline\red{Some\: important\: Formula:-}

SomeimportantFormula:−

\tt{\implies Perimeter\:_{(circle)}=2\:\pi\:r}⟹Perimeter

(circle)

=2πr

\tt{\implies Perimeter\:_{(semicircle)}=\pi\:r}⟹Perimeter

(semicircle)

=πr

\tt{\implies Area\:_{(circle)}=\pi\:r^2}⟹Area

(circle)

=πr

2

\tt{\implies Area\:_{(semicircle)}=\dfrac{\pi\:r^2}{2}}⟹Area

(semicircle)

=

2

πr

2

\tt{\implies Area\:_{(sector)}=\dfrac{\theta}{360}\times\:\pi\:r^2}⟹Area

(sector)

=

360

θ

×πr

2

\tt{\implies length\:_{(circle\:arc)}=\dfrac{\theta}{360}\times\:2\pi\:r}⟹length

(circlearc)

=

360

θ

Explanation: mark me as brainliest

Answered by Anonymous
12

Answer is 24!!!!

Hope this will help u

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