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Sauron [17]
4 years ago
10

Solve 5n=20 math right now please

Mathematics
2 answers:
maksim [4K]4 years ago
8 0

Answer:

n=4

Step-by-step explanation:

slega [8]4 years ago
7 0

Answer:

n=4

Step-by-step explanation:

5x4=20

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N+(-n) = 3+ 15<br><br> Solve
Len [333]

Answer:

no solutions


Step-by-step explanation:

n+(-n) = 3+ 15

n+-n = 0

0= 18

False

There are no solutions

7 0
3 years ago
Someone Help me please
OLEGan [10]

Answer:

-1/2

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
Lets see who is smart lol
Simora [160]

9514 1404 393

Answer:

  \square\quad{y=\dfrac{1}{2}x+4}\\\\\square\quad{\dfrac{y-5}{x-2}=\dfrac{1}{2}}

Step-by-step explanation:

The slope of a line is the same everywhere, so an equation can be written making use of that fact.

  \dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}\\\\\dfrac{y-5}{x-2}=\dfrac{7-5}{6-2}=\dfrac{2}{4}\\\\\boxed{\dfrac{y-5}{x-2}=\dfrac{1}{2}}

Cross-multiplying gives ...

  2(y -5) = x -2

  2y -10 = x -2  . . . eliminate parentheses

  2y = x +8 . . . . . . . add 10; next, divide by 2

  \boxed{y=\dfrac{1}{2}x+4}

__

These equations match choices B and D.

6 0
3 years ago
you get a 1 dollar coupon for every 3 salads you buy. What is the least number of salads you could buy to get 10 dollars in coup
miskamm [114]

Answer:

30 salads.

Step-by-step explanation:

If you can get a dollar coupon for 3 salads

You can just multiply it by 10 to get ten dollars for 30 salads

6 0
3 years ago
A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has i
goldfiish [28.3K]

Answer:

  • P(t) = 100·2.3^t
  • 529 after 2 hours
  • 441 per hour, rate of growth at 2 hours
  • 5.5 hours to reach 10,000

Step-by-step explanation:

It often works well to write an exponential expression as ...

   value = (initial value)×(growth factor)^(t/(growth period))

(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...

  P(t) = 100·2.3^t

__

(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours

__

(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t

  P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour

__

(d) We want to find t such that ...

  P(t) = 10000

  100·2.3^t = 10000 . . . substitute for P(t)

  2.3^t = 100 . . . . . . . . divide by 100

  t·log(2.3) = log(100)

  t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000

6 0
4 years ago
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