Over which part of the domain is the piecewise function defined as f(x) = 0.05x – 300?

Mathematics

2 answers:

Vesnalui [34]3 years ago

7 0

Answer:

B : For income 15,000 < x < 40,000

Step-by-step explanation:

I did the Edgenuity

Hope this helped :)

vovikov84 [41]3 years ago

3 0

Answer:it’s just b for future references

Step-by-step explanation:

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How to find (c,d,e)<br>Please do a step by step working.Thank you.

Natasha2012 [34]

(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
   $\frac{t + 1}{2}$ = y
Answer for (a): $g^{-1}(t)$ = $\frac{t + 1}{2}$

(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: $y =\frac{t - 3}{4}$

Answer for (b): $h^{-1}(t)$ = $\frac{t - 3}{4}$

(c)
$g^{-1} ( h^{-1}(t)) = g^{-1} (\frac{t - 3}{4})$
replace all t's in the $g^{-1}(t)$ equation with $\frac{t - 3}{4}$
$g^{-1} (\frac{t - 3}{4})$ = $\frac{ \frac{t-3}{4} + 1}{2}$
= $\frac{ \frac{t-3}{4} + \frac{4}{4}}{2}$ = $\frac{ \frac{t - 3 + 4}{4}}{2}$ = $\frac{ \frac{t + 1}{4}}{2} = \frac{t + 1}{8}$
Answer for (c): $g^{-1} ( h^{-1}(t))$ = $\frac{t + 1}{8}$

(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
t + 1 = 8y
$\frac{t + 1}{8}$ = y
Answer for (e): inverse of h(g(t)) = $\frac{t + 1}{8}$

8 0

3 years ago

Can you help me and give me the snow you got the answer right everything down

ExtremeBDS [4]

1. 2/100 as a percent is 2% and as a decimal it is 0.02

3. 9/10 as a percent is 90% and as a decimal it is 0.9

5. 48% as a decimal is 0.48 and as a fraction is 48/100

7. 0.04 as a percent is 4/100 and as a percent of is 4%

Hope I helped you

6 0

3 years ago

Find the equation of this line.<br>HELP

Mumz [18]

Answer:

Step-by-step explanation:

(5,4) ; (-3, -2)

$Slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{-2-4}{-3-5}\\\\=\frac{-6}{-8}\\\\=\frac{3}{4}$

(5,4) & m = (3/4)

y -y1 = m(x-x1)

y - 4 = (3/4)(x - 5)

$y-4=\frac{3}{4}x-\frac{3}{4}*5\\\\y-4=\frac{3}{4}x-\frac{15}{4}\\\\y=\frac{3}{4}x-\frac{15}{4}+4\\\\y=\frac{3}{4}x-\frac{15}{4}+\frac{4*4}{1*4}\\\\y=\frac{3}{4}x-\frac{15}{4}+\frac{16}{4}\\\\y=\frac{3}{4}x+\frac{1}{4}$