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irga5000 [103]
3 years ago
15

Find the domain of the rational function 9x/(x+5)(x+3)

Mathematics
1 answer:
xenn [34]3 years ago
7 0

Answer:

the domain is : D = ]-∞ , -5[ U ]-5 , -3[ U ]-3 , +∞[

Step-by-step explanation:

hello :

f(x) = 9x/(x+5)(x+3)

f exist for : (x+5)(x+3) ≠ 0

(x+5)(x+3)=0

x+5=0 or x+3=0  means : x= - 5 or x= -3

the domain is : D = ]-∞ , -5[ U ]-5 , -3[ U ]-3 , +∞[

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Righty in equation in standard form using integers Y +1 equals 2/3 parentheses X +3 parentheses
Korolek [52]

Answer:

The standard for of the equation would be 2x + 3y = 3

Step-by-step explanation:

To start, we simply need to solve for the constant.

y + 1 = 2/3(x + 3)

y + 1 = 2/3x + 2

2/3x + y + 1 = 2

2/3x + y = 1

Now we need to multiply all the numbers by the largest denominator. This will make all the numbers integers. In this case, that number would be 3.

2x + 3y = 3


4 0
3 years ago
- 16 + ( - 12 ) - ( - 40 )<br>​
Alex_Xolod [135]

Answer:

12

Step-by-step explanation:

Its 12 because when you do -16 + - 40.... then -12 you get =12 There you go

6 0
2 years ago
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Video games cost $23 each. Jake purchased 5 video games for $115.
11Alexandr11 [23.1K]
The equation for this is 23x5=115
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Why do I miss her? :/
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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

6 0
3 years ago
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