C-10-Q111mediumsingle_mcq

After rationalization, $\dfrac{20}{\sqrt{3}}$ becomes—

a $\dfrac{20\sqrt{3}}{3}$ m
b $10\sqrt{3}$ m
c $\dfrac{10}{\sqrt{3}}$ m
d $\dfrac{\sqrt{3}}{20}$ m

ব্যাখ্যা

Rationalizing means multiplying numerator and denominator by $\sqrt{3}$ : $\dfrac{20}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{20\sqrt{3}}{3}$ . This removes the surd from the denominator while keeping the value unchanged.

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q112Initial elevation of tower top is at with . Then equals—
C-10-Q113After moving m back to point where elevation is , equals—
C-10-Q114Combining the two equations gives
C-10-Q115Then equals—
C-10-Q116The approximate height of the tower is—
C-10-Q117From the initial point , elevation of the minar top is . The distance to the foot equals—
C-10-Q118After moving m towards the minar, the new distance to the foot equals—
C-10-Q119Combined equations give—
C-10-Q120The approximate height of the minar is—
C-10-Q121The exact (rationalized) form equals—

Chapter 10: Distance and Elevation — সব প্রশ্ন দেখুন (160টি) →

After rationalization, $\dfrac{20}{\sqrt{3}}$ becomes—

\dfrac{20\sqrt{3}}{3}

10\sqrt{3}

\dfrac{10}{\sqrt{3}}

\dfrac{\sqrt{3}}{20}

After rationalization, 203\dfrac{20}{\sqrt{3}}3​20​ becomes—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

After rationalization, 203\dfrac{20}{\sqrt{3}}3​20​ becomes—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

After rationalization, $\dfrac{20}{\sqrt{3}}$ becomes—

After rationalization, $\dfrac{20}{\sqrt{3}}$ becomes—