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mr_godi [17]
3 years ago
15

Solve nonhomogenous equation: x"+4x=cost

Mathematics
1 answer:
motikmotik3 years ago
6 0

Answer:

x(t) = Acos 2t + B sin 2t+\frac{cost}{3}

Step-by-step explanation:

Given is a non homogeneous second degree equation as

x"+4x=cost

Auxialary equation is

m^2+4 =0\\m = 2i, -2i

Hence general solution is

x = Acos 2t + B sin 2t

Particular integral is = \frac{cost}{D^2+4}

Since t has coefficient 1, we substitute

D^2 =-1\\PI = \frac{cost}{-1+4} =\frac{cost}{3}

Hence full solution is

x(t) = Acos 2t + B sin 2t+\frac{cost}{3}

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eimsori [14]
332/8= 41.5 but you cant make half of a blanket so itll be 41!
6 0
3 years ago
Rihanna must put a five dollar deposit down to buy a necklace. If P where presents the full price of the necklace which expressi
natta225 [31]

Answer:

<em>The expression is </em><em>P - 5</em>

Step-by-step explanation:

Rihanna already made a deposit of $5 to buy the necklace (given)

Full price of necklace = P

Let m = how much Rihanna has left to pay

Full price of necklace = amount paid + amount left to pay

= 5 + m

So we equate the two full necklace prices

5 + m = P

m = P - 5

Therefore, the amount Rihanna owes before tax = P - 5

6 0
3 years ago
You need to invest $1000 in a bank account and are give two options. The first option is to earn $50 every month you leave the m
garri49 [273]

Answer:

<h3><u>Option 1</u></h3>

Earn $50 every month.

  • Let x = number of months the money is left in the account
  • Let y = the amount in the account
  • Initial amount = $1,000

\implies y = 50x + 1000

This is a <u>linear function</u>.

<h3><u>Option 2</u></h3>

Earn 3% interest each month.

(Assuming the interest earned each month is <u>compounding interest</u>.)

  • Let x = number of months the money is left in the account
  • Let y = the amount in the account
  • Initial amount = $1,000

\implies y = 1000(1.03)^x

This is an <u>exponential function</u>.

<h3><u>Table of values</u></h3>

<u />

\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.

However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month.  To find this, graph the two functions and find the <u>point of intersection</u>.

From the attached graph, Account Option 1 accrues more until month 32.  From month 33, Account Option 2 accrues more in the account.

<h3><u>Conclusion</u></h3>

If the money is going to be invested for less than 33 months then Account Option 1 is the better choice.  However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.

3 0
2 years ago
Answer fast for Brainliest!
eimsori [14]
The term with the coeffecient is 20w
4 0
3 years ago
Simplify: -4x542x5
dimaraw [331]

multiply left to right:


-4 x 542 = -2,168

-2,168 x 5 = -10,840


answer: -10,840

6 0
2 years ago
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