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zimovet [89]
3 years ago
15

Write the following comparison as a ratio reduced to lowest terms 14 nickels to 15 dimes​

Mathematics
1 answer:
miskamm [114]3 years ago
7 0
14x5=70
15x10=150
Answer- 7:15
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5 divided by one is 0.2
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6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at
SOVA2 [1]

Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through  \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\

Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}

7 0
3 years ago
PLEASE HELPPPPP!!!!!!!!!!!!!!! SHOW WORK
larisa [96]

Answer:

  • 17. A. 180 in²
  • 18. B. 234 in²

Step-by-step explanation:

#17

<u>Total surface area is:</u>

  • A = 8(3*2 + 6*2) + 2(3*6) = 180 in²

Correct option is A

#18

<u>Total surface area is:</u>

  • A = 2(9*9) + 4(2*9) = 234 in²

Correct option is B

5 0
3 years ago
A play was attended by 984 persons. 75% of them were adults how many adults attend
Ierofanga [76]

Answer:

738

Step-by-step explanation:

984 ÷ 100 = 9.84

9.84 × 75 = 738

75/100

5 0
4 years ago
5. Write down the nth term of each of the following G.Ps
yawa3891 [41]

Answer:

1i) nth term = 12(-b/4)^(n-1)

ii) nth term = 3(-1/9)^(n-1)

2i) Sixth term= (-3b^5)/256

ii) Eight term = -1/1594323

Question:

The complete question as found on brainly (question-10746111):

Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.

i) 12,-3b,......sixth term

ii) 3,-1/3,..........;8th term?

Step-by-step explanation:

1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.

i) 12,-3b,......;sixth term

1st term = a= 12

r = common ratio = 2nd term /1st term

r = -3b/12 = -b/4

Then we would find the nth term

See attachment for details

ii) 3,-1/3,..........;8th term

1st term = a= 3

r = common ratio = 2nd term /1st term

r = (-⅓)/3 = (-⅓)(⅓) = -1/9

Then we would find the nth term

See attachment for details

2) we are to determine the sixth term in (i) and eight term in (ii)

See attachment for details

Sixth term= (-3b^5)/256

Eight term = -1/1594323

7 0
3 years ago
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