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maw [93]
3 years ago
13

An electric can opener is worth 36 dollars when it is new. After each year, it is worth half what it was the previous year. What

will it's worth be after 8 years? Round your answer to the nearest cent. Write the model.
Mathematics
1 answer:
Likurg_2 [28]3 years ago
4 0
So the equation is f(x)=36*(.5)^8, since your principal worth is 36 and it depreciates by 1/2 each year, which is 14 cents.
You might be interested in
Last month you purchased 6 individual songs and 1 albums for a total of $21.99.
Annette [7]

Answer:

Song =$1.50 Album =$12.99

Step-by-step explanation:

Lets assume Individual songs = x

albums=y

6x + y = 21.99-------- y= 21.99- 6x

4x + 3y= 44.97

Replace y in the second equation

4x+ 3(21.99-6x) = 44.97

4x +65.97- 18x = 44.97

14x= 21

x = 1.50

y= 21.99- (6*1.50)

= 12.99

7 0
3 years ago
Miguel bikes 23 km per hour and starts at mile 10. Gabby bikes 28 km per hour and starts at hour 0. Which system of linear equat
Goshia [24]

Answer:

Miguel: d = 23t + 10

Gabby: d = 28t

Step-by-step explanation:

We can write the equation for each case in slope-intercept form, d = mt + b

Where,

d = total distance

t = time

m (rate) = distance/hr

b = initial value

✔️ Miguel:

m = 23 km/hr

b = 10 mile

Equation: subtitle the values into d = mt + b

d = 23t + 10

✔️ Gabby:

m = 28 km/hr

b = 0 mile

Equation: subtitle the values into d = mt + b

d = 28t + 0

d = 28t

4 0
3 years ago
,,,,,,,,,hi help pls
aliya0001 [1]
Answer: slope is 0
to find the slope , you have to subtract y-es and divide them by x-es.(delta y/ delta x)
so,
(-6,-2) (-7,-2)
slope: -2-(-2)/-7-(-6)
slope:0/ -1
slope: 0
which looks on the graph like this:

4 0
3 years ago
2(x-7)+3=2x-10<br> This question has how many solutions
ycow [4]
I believe the answer is 0=1
8 0
2 years ago
Read 2 more answers
Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

7 0
3 years ago
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