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mote1985 [20]
2 years ago
11

Evaluate each expression. 1. 3x5+2x8+2

Mathematics
1 answer:
kirza4 [7]2 years ago
4 0

Answer:

1.23

2.60

3.45

4.37.524 (Im not incredibly posotive ab this one)

5.28

Step-by-step explanation:

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How many hours will it take a culture of bacteria to increase from 20 to 2000? Use k= 0.614
nikdorinn [45]

Answer:

it will take about 7.5 hours to increase from 20 to 2000

Step-by-step explanation:

y(t) = a  e^kt

20 = a e^ kt1

2000 = a e^ kt2

divide these equations

2000/20 = a e^ kt2/ a e^ kt1

when dividing the exponents subtract

100 =a/a e^(kt2-kt1)

factor out the k

100 = e ^ k(t2-t1)

take the natural log on each side

ln(100) = ln (e^ k(t2-t1)

ln (100) = k(t2-t1)

divide by k

ln(100)/k = (t2-t1)

we know k=.614

ln(100)/.614 = t2-t1

7.500277 = t2-t1

it will take about 7.5 hours to increase from 20 to 2000

5 0
3 years ago
∇f = λ∇g gives us the equations 4xy = 4λx, 2x2 = 8λy. The first equation implies that x = Incorrect: Your answer is incorrect. o
Vlad1618 [11]

Answer:

The answer is correct

Step-by-step explanation:

6 0
3 years ago
What is the end behavior of the polynomial function?
valina [46]
No se pero me gustaría alludar
3 0
2 years ago
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A rectangular garden measures 25ft by 38ft. Surrounding (and bordering) the garden is a path 2ft wide. Find the area of this pat
lesya [120]

So as the path adds an extra 2 feet to each side I would start this by adding two feet to each part (so 25+2 x 38+2) which leaves you with 27ft by 40ft. Multiply to find this total area, which results in 1080. Now, you want to remove the actual area of the garden itself as this is not part of the path. So 25x38=950. Subtract 1080 by 950. This gives you just the area of the gardens path which would be 130 feet^{2}

4 0
3 years ago
Read 2 more answers
If 75% of the students in Toby's grade voted, how many students are in Toby's grade?
kupik [55]

Answer:

HOW SHOULD WE KNOW WE DONT HAVE ANY INFORMATION EXCEPT 75% AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Step-by-step explanation:

3 0
3 years ago
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