LAST QUESTION I think it's 12/5, am I right?

Mathematics

1 answer:

Anon25 [30]3 years ago

8 0

Yes u are right hope this helps

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(1 ÷ x- 1)+(2÷ x+2)=(3÷2) solve and check for extraneous solutions

Sliva [168]

$\frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2}$
$\frac{1(x+2)}{(x-1)(x+2)}+ \frac{2(x-1)}{(x+2)(x-1)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}$
$\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}$
$\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}$
$\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}$
$\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0$
$\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0$

this will be continued

5 0

4 years ago

As pointed out, you lose if the pile has 1, 6, 11 or 16 sticks. How many sticks are needed in the next two piles for you to lose

muminat

Answer:

21 and 26 totaling 47sticks

Step-by-step explanation:

Th number of sticks form an arithmetic progression

1, 6, 11, 16...

The nth term of an arithmetic progression is expressed as;

Tn = a+(n-1)d

a is the first term = 1