Rohit said ", I hit a century yesterday" convert it in to reported speechplz give ans it's very argent​

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. hope it helps you (◕ᴗ◕✿)

Answer:

There are three major categories based on which computers can be classified. These are: Based on Size. Based on Purpose.

...

Answer: (4) Laptop.

Computer Networks Microsoft Office

Microsoft Windows Web Browsers

Computer Input and Output Devices

Answer:

There are three types of the computer on the basis of work. They are analog, digital, and hybrid computer.

Analog - thermometer, voltmeter, etc

Digital - Personal computer, mobile phone

Hybrid - ultrasound machine, monitoring machine

Answer:

3,1,-1,-3,-5,-7,-9,-11,-13

Answer:

[tex]\qquad\qquad\boxed{ \sf{ \: \: \bf \:\dfrac{tan(A + B)}{sinA} = \dfrac{3}{2} \: }} \\ \\ [/tex]

Step-by-step explanation:

Given that,

[tex]\sf \:sinB = \dfrac{1}{5}sin(2A + B) \\ \\ [/tex]

can be rewritten as

[tex]\sf \:\dfrac{sin(2A + B)}{sinB} = 5 \\ \\ [/tex]

[tex]\sf \:\dfrac{sin(2A + B)}{sinB} = \dfrac{5}{1} \\ \\ [/tex]

On applying Componendo and Dividendo, we get

[tex]\sf \:\dfrac{sin(2A + B) + sinB}{sin(2A + B) - sinB} = \dfrac{5 + 1}{5 - 1} \\ \\ [/tex]

can be further rewritten as

[tex]\sf \:\dfrac{2sin\bigg( \dfrac{2A + B + B}{2} \bigg)cos\bigg( \dfrac{2A + B - B}{2} \bigg) }{2cos\bigg( \dfrac{2A + B + B}{2} \bigg)sin\bigg( \dfrac{2A + B - B}{2} \bigg)} = \dfrac{6}{4} \\ \\ [/tex]

[tex]\sf \:\dfrac{sin\bigg( \dfrac{2A + 2B}{2} \bigg)cos\bigg( \dfrac{2A}{2} \bigg) }{cos\bigg( \dfrac{2A + 2B}{2} \bigg)sin\bigg( \dfrac{2A }{2} \bigg)} = \dfrac{3}{2} \\ \\ [/tex]

[tex]\sf \:\dfrac{sin(A + B)cosA}{cos(A + B)sinA} = \dfrac{3}{2} \\ \\ [/tex]

[tex]\sf\implies \: \sf \:\dfrac{tan(A + B)}{sinA} = \dfrac{3}{2} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used

[tex]\sf \:sinx +siny = 2sin\bigg( \dfrac{x + y}{2} \bigg)cos\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]

[tex]\sf \:sinx - siny = 2cos\bigg( \dfrac{x + y}{2} \bigg)sin\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\sf \:cosx +cosy = 2cos\bigg( \dfrac{x + y}{2} \bigg)cos\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]

[tex]\sf \:cosx - cosy \: = \: - \: 2sin\bigg( \dfrac{x + y}{2} \bigg)sin\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]

[tex]\large{\bf{Question:}}[/tex]

If sin B = ⅕sin(2A + B), then tan(A+B)/tan A =

Solution:

[tex]\qquad\bf{sinB= \dfrac{sin(2A+B)}{5}} \\ \\[/tex]

[tex]\qquad\qquad\sf{⇒ \dfrac{sin(2A+B)}{sinB} \:= \dfrac{5}{1}} \\ \\[/tex]

Using componendo and dividendo,

we get

[tex]\qquad\sf{\dfrac{sin(2A+B)+sinB}{sin(2A+B)−sinB} = \dfrac{5+1}{5-1}} \\ \\[/tex]

[tex]\qquad\sf{⇒ \dfrac{2sin(A+B)cosA}{2cos(A+B)sinA} = \dfrac{6}{2}} \\ \\[/tex]

[tex]\qquad\sf{⇒ \dfrac{tan(A+B)}{tanA} = \dfrac{3}{2}} \\ \\[/tex]