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elena-14-01-66 [18.8K]
3 years ago
7

W+2/3=2 5/6 I can't figure out the answer

Mathematics
1 answer:
insens350 [35]3 years ago
8 0
If you're solving for W, your answer is w = 2 1/6

First, we would simplify our fractions:
 w + 2/3 = 2 5/6 =
w + 2/3 = 17/6

Lastly, we subtract 2/3 on both sides of the equation:
w + 2/3 - 2/3 = 17/6 - 2/3 =
w = 13/6 or w = 2 1/6
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1. Write the equation of the piece-wise function graphed below​.
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Problem 4

<h3>Answer:</h3>

f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}

------------------

Work Shown:

The left line goes through (-2,0) and (0,4)

The slope of this line is

m = (y2-y1)/(x2-x1)

m = (4-0)/(0-(-2))

m = (4-0)/(0+2)

m = 4/2

m = 2

The y intercept is b = 4

Since m = 2 and b = 4, this means y = mx+b turns into y = 2x+4. This portion is only done when x < 1. Note the open circle at the endpoint of this portion. So we do not include x = 1 as part of this piece.

---

The line on the right side goes through (1,-2) and (2,-1)

Slope

m = (y2-y1)/(x2-x1)

m = (-1-(-2))/(2-1)

m = (-1+2)/(2-1)

m = 1/1

m = 1

The y intercept is b = -3. You can see this if you extend the line until it crosses the y axis.

Alternatively, plug in (x,y) = (1,-2) and m = 1 into y = mx+b to find that b = -3

So y = mx+b turns into y = 1x+(-3) or just y = x-3

We combine both parts to end up with f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}

This is only graphed when x \ge 1 (note the closed or filled in circle for the endpoint of this portion).

===================================================

Problem 5

Answer:

<h3>f(x) = \frac{1}{2}|x+3| is the absolute value function</h3><h3>while this is the piecewise function</h3>

f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\

------------------

Work Shown:

y = |x| .... parent function

y = |x+3| ... shift 3 units to the left

y = (1/2)*|x+3| .... vertically compress by factor of 1/2

f(x) = (1/2)*|x+3|

------

Break that down into a piecewise function

when x < -3, then y = -(1/2)(x+3)

when x \ge -3, then y = (1/2)(x+3)

I'm using the rule that y = |x| turns into y = -x when x < 0 and y = x when x \ge 0

So that is how we get f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\as the piecewise function.

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Answer:

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