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kow [346]
3 years ago
11

Given the function f(x)=1/2x+8 find x so that, f(x)=10 please help im stumped

Mathematics
1 answer:
Crank3 years ago
3 0

Step-by-step explanation:

plug y value :

10=1/2x+8

1/2x=2

x=4

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Read 2 more answers
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Consider the functions given below. SEE F
Wewaii [24]

Answer:

1. P(x) ÷ Q(x)---> \frac{-3x + 2}{3(3x - 1)}

2. P(x) + Q(x)---> \frac{2(6x - 1)}{(3x - 1)(-3x + 2)}

3.  P(x) - Q(x)---> \frac{-2(12x - 5)}{(3x - 1)(-3x + 2)}

4. P(x)*Q(x) --> \frac{12}{(3x - 1)(-3x + 2)}

Step-by-step explanation:

Given that:

1. P(x) = \frac{2}{3x - 1}

Q(x) = \frac{6}{-3x + 2}

Thus,

P(x) ÷ Q(x) = \frac{2}{3x - 1} ÷ \frac{6}{-3x + 2}

Flip the 2nd function, Q(x), upside down to change the process to multiplication.

\frac{2}{3x - 1}*\frac{-3x + 2}{6}

\frac{2(-3x + 2)}{6(3x - 1)}

= \frac{-3x + 2}{3(3x - 1)}

2. P(x) + Q(x) = \frac{2}{3x - 1} + \frac{6}{-3x + 2}

Make both expressions as a single fraction by finding, the common denominator, divide the common denominator by each denominator, and then multiply by the numerator. You'd have the following below:

\frac{2(-3x + 2) + 6(3x - 1)}{(3x - 1)(-3x + 2)}

\frac{-6x + 4 + 18x - 6}{(3x - 1)(-3x + 2)}

\frac{-6x + 18x + 4 - 6}{(3x - 1)(-3x + 2)}

\frac{12x - 2}{(3x - 1)(-3x + 2)}

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

3. P(x) - Q(x) = \frac{2}{3x - 1} - \frac{6}{-3x + 2}

\frac{2(-3x + 2) - 6(3x - 1)}{(3x - 1)(-3x + 2)}

\frac{-6x + 4 - 18x + 6}{(3x - 1)(-3x + 2)}

\frac{-6x - 18x + 4 + 6}{(3x - 1)(-3x + 2)}

\frac{-24x + 10}{(3x - 1)(-3x + 2)}

= \frac{-2(12x - 5}{(3x - 1)(-3x + 2)}

4. P(x)*Q(x) = \frac{2}{3x - 1}* \frac{6}{-3x + 2}

P(x)*Q(x) = \frac{2*6}{(3x - 1)(-3x + 2)}

P(x)*Q(x) = \frac{12}{(3x - 1)(-3x + 2)}

4 0
4 years ago
Can someone please help me out pleaseeeee !!!!!
zysi [14]

Answer:

Step-by-step explanation:

0: -2

1: -4

2: -6

4 0
3 years ago
One way to show that a statement is not a good definition is to find a
guapka [62]

counterexaple

i hope it helps

4 0
3 years ago
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