C-10-Q050mediumsingle_mcq

After rationalizing, $\dfrac{75}{\sqrt{3}}$ becomes—

a $75\sqrt{3}$ m
b $25\sqrt{3}$ m
c $15\sqrt{3}$ m
d $\dfrac{25}{\sqrt{3}}$ m

ব্যাখ্যা

Multiplying numerator and denominator by $\sqrt{3}$ gives $\dfrac{75\sqrt{3}}{3} = 25\sqrt{3}$ m. So the tower's height is $25\sqrt{3} \approx 43.3$ m.

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q051The approximate height of the tower in Example 1 is—
C-10-Q052To rationalize , the numerator and denominator are multiplied by—
C-10-Q053In Example 2, with tree m, foot distance , and , equals—
C-10-Q054The distance satisfies—
C-10-Q055simplifies to—
C-10-Q056The approximate ground distance of the point in Example 2 is—
C-10-Q057The relation tells us that the side opposite the angle is times the—
C-10-Q058With the same height m, if the elevation were instead of , the foot distance would be—
C-10-Q059A ladder of length m leans against a wall making with the ground; with m and wall height , equals—
C-10-Q060Therefore the wall height equals—

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

After rationalizing, $\dfrac{75}{\sqrt{3}}$ becomes—

75\sqrt{3}

25\sqrt{3}

15\sqrt{3}

\dfrac{25}{\sqrt{3}}

ব্যাখ্যা

Multiplying numerator and denominator by $\sqrt{3}$ gives $\dfrac{75\sqrt{3}}{3} = 25\sqrt{3}$ m. So the tower's height is $25\sqrt{3} \approx 43.3$ m.

After rationalizing, 753\dfrac{75}{\sqrt{3}}3​75​ becomes—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

After rationalizing, 753\dfrac{75}{\sqrt{3}}3​75​ becomes—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

After rationalizing, $\dfrac{75}{\sqrt{3}}$ becomes—

After rationalizing, $\dfrac{75}{\sqrt{3}}$ becomes—