So
lets say we have
a/b=2a/2b
we know that if we invert one and multiply it by the other (divide them), we get 1 because a/a=1 where a=a

so
2/3 and 4/6 are equivelent because if you divide them we get 12/12=1

2/3 and 8/12 are equivilent because if you divide them we get 24/24=1

and sinde 2/3=4/6 and 2/3=8/12, 2/3=4/6=8/12

they are equivlent

6 0

3 years ago

A^3+a+a^2+1 2a^2+2ab+ab^2+b^3

Vikentia [17]

$\frac{a^3+a+a^2+1}{a^3+a^2+ab^2+b^2} \cdot \frac{2a^2+2ab+ab^2+b^3}{a^3+a+a^2b+b} =\frac{a(a^2+1)+1(a^2+1)}{a^2(a+1)+b^2(a+1)} \cdot \frac{2a(a+b)+b^2(a+b)}{a(a^2+1)+b(a^2+1)} =\\\\=\frac{(a^2+1)(a+1)}{(a+1)(a^2+b^2)} \cdot \frac{(a+b)(2a+b^2)}{(a^2+1)(a+b)} = \frac{2a+b^2}{a^2+b^2}$

6 0

3 years ago

If tan tetha =8/15,find the value of sin tetha+cos tetha all divided by cos tetha (1-cos tetha)

Dahasolnce [82]

Answer: 195.5 or 12 7/32

Step-by-step explanation:

There is no letter tetha in the table so I use α instead. However it is not sence to final result.

The expression is:

(sinα+cosα)/(cosα*(1-cosα))

Lets divide the nominator and denominator by cosα

(sinα/cosα+cosα/cosα)/(cosα*(1-cosα)/cosα)= (tanα+1)/(1-cosα)=

=(8/15+1)/(1-cosα)= 23/(15*(1-cosα)) (1)

As known cos²α=1-sin²α (divide by cos²α both sides of equation)

cos²a/cos²α=1/cos²α-sin²α/cos²α

1=1/cos²α-tg²α

1/cos²α=1+tg²α

cos²α=1/(1+tg²α)

cosα=sqrt(1/(1+tg²α))= +-sqrt(1/(1+64/225))=+-sqrt(225/(225+64))=

=+-sqrt(225/289)=+-15/17 (2)

Substitute in (1) cosα by (2):

1st use cosα=15/17

1) 23/(15*(1-cosα)) =23/(15*(1-15/17))= 23*17/2=195.5

2-nd use cosα=-15/17

2)23/(15*(1-cosα)) =23/(15*(1+15/17))= 23*17/32=12 7/32

7 0

3 years ago