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lord [1]
1 year ago
15

One function has an equation in slope-intercept form: y = x + 5. Another function has an equation in standard form: y + x = 5. E

xplain what must be different about the properties of the functions. See if you can determine the differences without converting the equation to the same form.
Mathematics
1 answer:
vlada-n [284]1 year ago
7 0

Without converting the equations to the same form, the property that must be different in the functions is the slope

<h3>How to determine the difference in the properties of the functions?</h3>

From the question, the equations are given as

y = x + 5

y + x = 5


From the question, we understand that:

The equations must not be converted to the same form before the question is solved

The equation of a linear function is represented as

y = mx + c

Where m represents the slope and c represents the y-intercept

When the equation y = mx + c is compared to y = x + 5, we have

Slope, m = 1

y-intercept, c = 5

The equation y = mx + c can be rewritten as

y - mx = c

When the equation y - mx = c is compared to y + x = 5, we have

Slope, m = -1

y-intercept, c = 5

By comparing the properties of the functions, we have

  • The functions have the same y-intercept of 5
  • The functions have the different slopes of 1 and -1

Hence, the different properties of the functions are their slopes

Read more about linear functions at

brainly.com/question/15602982

#SPJ1

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Step-by-step explanation:

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EXPLANATION

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5 0
3 years ago
The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth
Art [367]

Answer:

a) N_0=250\; k=0.15

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies t=0. And N(0)=N_0. The process is:

N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250

b) After 2 days means t=48. So, we just replace and operate:

N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}

c) N(t_1)=4000; \;t_1=?

N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\  \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h

d) t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)

N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h

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