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3 years ago

The Bayside Bugle charges by the word to run automotive ads. The newspaper

Mathematics

1 answer:

11111nata11111 [884]3 years ago

8 0

Answer:

$20.45

Step-by-step explanation:

Since the Bayside Bugle charges $18 for the first 20 words and there is more than 20 words in the ad you take 27-20 and you get 7 additional words.

Now we know that we already have to pay $18 for the first $20 words.

Since we have 7 additional words we multiply that by 0.35 cents and you get $2.45.

Next we add up our two numbers (2.45+18) we get the total of $20.45 fro a 27-word ad.

I hope this helps :)

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The differential equation of a certain system is 20⁢y′⁢′+c⁢y′+80⁢y=0

Sholpan [36]

Answer:

c=80

Step-by-step explanation:

Based on my reading the critical damping occurs when the discriminant of the quadratic characteristic equation is 0.

So let's see that characteristic equation:

20⁢r^2+c⁢r+80⁢=0

The discriminant can be found by calculating b^2-4aC of ar^2+br+C=0.

a=20

b=c

C=80

c^2-4(20)(80)

We want this to be 0.

c^2-4(20)(80)=0

Simplify:

c^2-6400=0

Add 6400 on both sides:

c^2=6400

Take square root of both sides:

c=80 or c=-80

Based on further reading damping equations in form

a⁢y′⁢′+b⁢y′+C⁢y=0

should have positive coefficients with b also having the possibility of being zero.

5 0

3 years ago

Which equation demonstrates the associative property of multiplication

kherson [118]

Answer:

from what I remember ide say A

3 0

3 years ago

Add x^3 - 4x^2 + 1 to 3x^2 + x <br> Show Your work!

Sladkaya [172]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: (x {}^{3} - 4x {}^{2} + 1) + (3 {x}^{2} + x)$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 1 + 3 {x}^{2} + x$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 3x {}^{2} + x + 1$

$\qquad \sf \dashrightarrow \: {x}^{3} - {x}^{2} + x + 1$

6 0

2 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

Explanation please <br><br>thank you

GrogVix [38]

300 represents the initial value or the value that you started with.
1.15 is the rate that the function increases by over t (time)

8 0

3 years ago