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padilas [110]
3 years ago
8

Please help me Simplify 5²- 2²

Mathematics
2 answers:
IgorC [24]3 years ago
6 0

Answer:

21

Step-by-step explanation:

5x5= 25

2x2= 4

25-4= 21

Kobotan [32]3 years ago
3 0

Answer:

21

Step-by-step explanation:

Simplify 5^2 to 25

25-2^2

Simplify 2^2 to 4

25-4

Answer

21



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Find the equation of the line given the following
Oduvanchick [21]

Answer:

1. y = -10x + 6

2. y = -x - 6

3. y = 13x + 110

4. y = -5x + 47

5. y = 7/4 x -7

Step-by-step explanation:

2. m = - 1 and Y-intercept: -20 - (- 1)(14) = -6

3. m = 13 and Y-intercept: -7 - (13)(- 9) = 110

4. (10,-3) and (7,12)

Slope = (12 - - 3)(7- 10) = 15/-3 = -5

y- intercept = 12 - (-5)(7) = 47

5. (4,0) and (0, -7)

Slope: (-7-0)/(0-4) = -7/-4 = 7/4

y-intercept: -7 - 7/4)(0) = -7

7 0
3 years ago
When u are doing math and u have a question with pi in it don't u use "3.14" or do u use "3.141" or is it "3.14159?
Tju [1.3M]
Usually, I use 3.14.
7 0
4 years ago
Read 2 more answers
What is the mean of integers<br> -16,-27,21,-19,14,-3<br> -
Lelu [443]

Answer:

-5

Step-by-step explanation:

-16+(-27)+21+(-19)+14+(-3)/6

-30/6

=-5

3 0
3 years ago
I don’t understand this pls help me please :(
Irina-Kira [14]
The answer is A) C=0.45p.
To find the cost you must multiply the cost per pound by the number of pounds purchased to get the total cost. That's what choice A is.
7 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

6 0
3 years ago
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