Question

Jordan is cutting a 2 \text{ m}2 m2, start text, space, m, end text by 1\dfrac14\text{ m}1
4
1
​
m1, start fraction, 1, divided by, 4, end fraction, start text, space, m, end text piece of rectangular paper into two pieces along its diagonal.

Answer 1

please write the question clearly!!

ANSWER:

•  Radius = 2.5 cm 

Given:

Radius of spherical ball is 3 cm.

Radii of new spherical balls are 1.5 cm and 2 cm.

To Find:

Radius of third spherical ball ?

Solution :

 Let the radius of third spherical ball be x cm.

If something is melted and recasted into another thing then their volumes will be equal. In short

Volume of 1st thing = Volume of second one.

➯ Let's see here

Volume of big spherical ball will be equal to the sum of volumes of that three small spherical balls.

As we know that

★ Volume of Sphere = 4/3πr³ ★

[ Taking big spherical ball ]

Radius = 3 cm

⟹ Volume = 4/3 × π × (3)³

⟹ 4π/3 × 27

Volume we got = 4π/3 × 27 cm³

[ Taking 3 small spherical balls ]

Radius of first ball (r¹) = 1.5 cm

For second (R) = 2 cm

For third (x) = x cm

Volume = 4/3 × π( sum cubes of radii)

⟹ Volume = 4/3 × π(1.5³ + 2³ + r³)

⟹ 4π/3 (3.375 + 8 + x³)

⟹ 4π/3 ( 11.375 + x³)

Volume we got = 4π/3 (11.375 + x³) cm³

A/q

First volume = Second volume

➮ 4π/3 × 27 = 4π/3 (11.375 + x³)

➮ 27 = 11.375 + x³

➮ 27 – 11.375 = x³

➮ 15.625 = x³

➮ 15625/1000 = x³

➮ 3125/200 = 625/40 = 125/8 = x³

➮ ³√125/8 = x³

➮ 5/2 = x²

➮ 2.5 cm = x

Hence, the measure of radius of third spherical ball is 2.5 cm.

Answer 2

Step-by-step explanation:

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Trigonometric table :-

$\begin{gathered}\begin{gathered}\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}\end{gathered}\end{gathered}$