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3 years ago

I need a quick good conclusion thank you

Mathematics

1 answer:

Debora [2.8K]3 years ago

8 0

You'll notice that 36's prime factors had 2 even, and none other had an even number of even prime factors. Therefore, if you have an even number of even prime factors, your square root will be rational given the data

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Convert this radical form to exponential form.

Elden [556K]

Answer:

I think it's 7.93725

I'm not sure if I did this correctly

3 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago

1. Find the LCM of the given numbers. 4 and 7

GuDViN [60]

Answer:

The LCM of (4,7) is 28

Step-by-step explanation:

7 0

2 years ago

PLEASE HELP

Nady [450]

The frequency for rolling a 3 would be 5 out of 100

6 0

3 years ago

15-3(2+6 times -3) What would this simplify to?

pickupchik [31]

-45 because of PEMDAS

8 0

3 years ago

Read 2 more answers