Question

Algebra 2 Help @Ranga Can you help please, my math angel? Will attach the problem.

Answer 1

I think it's c

U could have 2x(x+1)

It'll be hard for b becuz what would u multiply for the first fraction to get 2x+1 from x+1

Answer 2

Okay, so, um, my grandpa and I wanna know how do you find the LCD of a regular fraction??

Answer 3

1/2 + 2/3. Can you find the LCD for this problem?

Answer 4

No idea. Is it 2? But, could you help us figure out the actual problem? We keep reworking it, but nothing we get matches the answers given. .-.

Answer 5

Oh, he just got a matching answer! Thanks anyway! :)

Answer 6

The LCD is the smallest number that the denominator can be divided into by both numbers and result in a whole number. 1/2 + 2/3 The denominator are 2 and 3. The LCD would be 6. 2 x3=6. Thus the problem would now be 3/6 + 4/6 + 7/6

Answer 7

2x(x+1)
All three denominators: x, 2x and x+1 can divide into 2x(x+1).

Answer 8

LCD is same as finding LCM for a bunch of numbers. When the numbers are in the denominator of a fraction we call it LCD.

LCM of 2 and 3 is 6. Both 2 and 3 will divide evenly into 6. But how about 12? 2 and 3 will divide evenly into 12 also but it is not the LEAST number. There is a smaller number than 12, which is 6 and that is the least common denominator.

LCM of 2 and 4 is 4.

LCM of 2, 3, 5 is: 30.

LCM of 4, 5, 6, 7 is: 420.

Answer 9

Oh my gosh, I forgot to close this question. We got it all figured out a little while ago. Can you help with another question, though??

Answer 10

Okay.

Answer 11

Thank you! Okay, here's the question:
Amanda tells you that when variables are in the denominator, the equation becomes unsolvable. "There is a value for x that makes the denominator zero, and you can't divide by zer," Amanda explains. Using complete sentences, demonstrate to Amanda how the equation is still solvable.

Answer 12

When there are variable expressions in the denominator we have to make sure the variable does not taken on a value that would make the denominator zero.

For example, in the expression 15 / (x-3), x cannot be 3.

But we can still solve the equation, but if any of the solution has x = 3 (as in this example) we discard that solution.

Answer 13

Example: Solve: (x^2 - 4) / (x + 2) = 5.

The denominator has the expression (x+2) which cannot be zero. Therefore, x cannot take on the value -2.

But we can still solve this equation. The numerator can be factored as: (x+2)(x-2)
(x^2 - 4) / (x + 2) = 5 becomes
(x+2)(x-2) / (x+2) = 5 (the (x+2) will cancel out)
x - 2 = 5
x = 7

Answer 14

Since only x = -2 is not allowed, the solution x = 7 is a valid solution to the equation:
(x^2 - 4) / (x + 2) = 5.

Answer 15

In fact you can verify that x = 7 is a valid solution by putting it back in the original equation:
(x^2 - 4) / (x + 2) at x = 7 is
(7^2 - 4) / (7 + 2) = 45 / 9 = 5 which is the right hand side of the original equation.

Answer 16

Sweet, thanks Ranga! You're a real life saver, ya know? Can you help me with a few more? They look like just explaining stuff.

Answer 17

I can do one more now after which I have to log off to attend to other matters.

Answer 18

Okay, cool, thanks! I'll type the next one up, since I can't copy/paste. ^^;;;

Answer 19

Here's the question:
Pablo was graphing a function and noticed that at certain points, the graph reaches invisible lines the graph will never cross. Explain to Pablo what the two types of invisible lines are and how to predict them. You may create your own example ito aid in your reasoning. Use complete sentences.

Answer 20

The invisible lines are called asymptotes.
Mainly there are two asymptotes: horizontal and vertical.
For example the function f(x) = 1 / (x-3) has a vertical asymptote at x = 3.
Since the the denominator can never be zero, x can never be 3. But x can get really, really close to 3 from either side but it can never get to 3. Examples: 2.9999999 and 3.0000001.
The y value will get very large as x approaches 3. And x = 3 is the invisible line that the graph won't cross. This is a vertical asymptote.

There is horizontal asymptote too.