Solve absolute value equation. Check for extraneous solution. |2x-3|=2x-1. Post it into the drop box for the unit 2 review.

1 answer:

6 0

let's keep in mind that, an absolute value expression is in effect a piecewise one, since it has a ± versions of it.

$\bf |2x-3|=2x-1\implies \begin{cases}+(2x-3)=2x-1\\-(2x-3)=2x-1\end{cases}\\\\[-0.35em]~\dotfill\\\\+(2x-3)=2x-1\implies 2x-3=2x-1\implies -3 \ne -1\impliedby extraneous\\\\[-0.35em]~\dotfill\\\\-(2x-3)=2x-1\implies 2x-3=-2x+1\implies 4x=4\\\\\\x=\cfrac{4}{4}\implies x=1$

You might be interested in

Which expression is equal to 5, ? A 542 B. 51 Cw D. 254

nirvana33 [79]

Answer:

don't understand the full question

3 0

3 years ago

Verify that a+b=b+a by taking a=6 and b=3/6

Klio2033 [76]

Answer:

Step-by-step explanation:

substitute all the numbers in

6+3/6+3/6+6=13

4 0

3 years ago

The weight of oranges growing in an orchard is normally distributed with a mean weight of 6.5 oz. and a standard deviation of 1.

sergey [27]

Add 3 standard deviations above and below the mean to get the range in which 99.7% of the data in a normal distribution will fall
6.5 + 4.5 = 11
6.5 - 4.5 = 2
So 2 to 11 ounces would be the interval

6 0

3 years ago

10 Pts Multiplying fractions: 5 1 — x — = 6 2

Flauer [41]

Multiply the numerators together. Multiply the denominators together. Then reduce if necessary.

$\dfrac{first~numerator}{first~denominator} \times \dfrac{second~numerator}{second~denominator} = \dfrac{(first~numerator)\times(second~numerator)}{(first~denominator)\times(second~denominator)}$

$\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$

$\dfrac{5}{6} \times \dfrac{1}{2} = \dfrac{5 \times 1}{6 \times 2} = \dfrac{5}{12}$

6 0

4 years ago

Read 2 more answers

What is the mode? 0-0-2-1-3-2-1-0

exis [7]

The mode is 0 since it is the most frequent.

4 0

3 years ago

Read 2 more answers