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2 years ago

Someone help please I'm really stuck on these questions and I need help please I need to bring this in tomorrow please help

Mathematics

1 answer:

AysviL [449]2 years ago

7 0

1) 1
2) PA: between 2 and 3.
2) PB: Unsure about it.

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3 years ago

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Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.

Blizzard [7]

Answer:

The answer is

$y = 3$

$y = - 4$

Step-by-step explanation:

We must find a solution where

$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

$\frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1}$

Which equals

$\frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)}$

Add the fractions

$\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Simplify the right side by multiplying the fraction

$\frac{6y}{(y + 1)(y + 2)}$

Set both fractions equal to each other

$\frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}$

Since the denomiator are equal, we must set the numerator equal to each other

$6y = {y}^{2} + 7y - 12$

$= {y}^{2} + y - 12$

$(y + 4)(y - 3)$

$y = - 4$

$y = 3$

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2 years ago

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Find the domain of fg

vazorg [7]

We have the following functions:
f (x) = x ^ 2 + 1
g (x) = 1 / x
Multiplying we have:
(f * g) (x) = (x ^ 2 + 1) * (1 / x)
Rewriting:
(f * g) (x) = ((x ^ 2 + 1) / x)
Therefore, the domain of the function is given by all the values of x that do not make zero the denominator.
We have then:
All reals except number 0
Answer:
b. all real numbers, except 0

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