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umka2103 [35]
3 years ago
12

(2x+1) -2 = (2x-1) -2 solve

Mathematics
2 answers:
anastassius [24]3 years ago
7 0

Answer:

Ok

Step-by-step explanation:

-2 Cancel out from both sides

(2x + 1) = (2x - 1)

Kai the answer is 0= -2

ElenaW [278]3 years ago
6 0

Answer:

There is no solution.

Step-by-step explanation:

2x+1=2x-1

2x-2x=-1-1

0=-2

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One hundred people in a supermarket survey
tiny-mole [99]

Answer:

13%

Step-by-step explanation:

So basically the people who had a pet in the store were: 47, 33, and 7 out of 100. Now we add up the numbers of the people who have pets which in total, is 87, out of 100. But that's not all. Those are the people who have pets to find the people who didn't have pets, we subtract 87 from 100 to get 13, or 13% as our answer.

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3 0
3 years ago
How many times dose 1/3 go into 3/4
Varvara68 [4.7K]

Answer:

2 times

Step-by-step explanation:

1/3=.33 and 3/4=.75,so .33*2 is .66,which is still smaller than .75

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4 0
3 years ago
Read 2 more answers
Simplify -8/2 ÷ 6/-3<br> Answer choices:<br> A: -4<br> B: 2<br> C:-2<br> D: 4
motikmotik

Answer:

2

Step-by-step explanation:

-8/2=-4

6/-3=-2

-4/-2=2

7 0
3 years ago
Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
3 years ago
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