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Natali5045456 [20]
3 years ago
9

2.) What amount presently must be invested earning 5.25% compounded continuously

Mathematics
2 answers:
julsineya [31]3 years ago
5 0

The amount A resulting from a principal amount P being invested at rate r compounded continuously for time t is given by

... A = P·e^(rt)


FIll in your given values and solve for P.

... 25000 = P·e^(0.0525·12) = P·e^0.63

... P = 25000/e^0.63 ≈ 13314.80 . . . . . divide by the coefficient of P


The amount that must be invested is $13,314.80.

andre [41]3 years ago
5 0

An initial investment (P) compounded continuously with a rate of interest (r) in time (t) will grow to amount (Q) is given by:

Q = P * e^(rt)

Q=25000, r=0.0525, t=12

25000 = P * e^(0.0525*12)

1.8776P = 25000

P = 13314.8

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113.0ft squared

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nadya68 [22]

(a) There are 1/4 cups of flour per serving

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(b) There are aprox. .18 cups of sugar per serving.

<u>Work:</u> 9/12+4/12=13/12 n then divide 13/12 by 6

(c) You need 2 1/4 cups of flour to make 9 servings

<u>Work:</u> 1.5/2=.75        1.5+.75=2.25=2 1/4 cups

(d) Someone would need to know the amount of sugar per serving to make the recipe accurate.

<em>Question 2</em>

(a) You're account is at $10 dollars

<u>Work:</u> $5x5 (days)=$25. add 25 to -15 and you have $10 dollars in you're account.

(b) 5 days.

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6 0
2 years ago
Simplify:<br> 2x²(4x³-3x²+6x)
Alinara [238K]

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244x²

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Answer:

<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.

Step-by-step explanation:

We need to form a LPP( Linear Programming Problem) for the following problem and solve it to get the maximize the profit.

Let x denote the number of men's shoe and y denote the number of female shoes.

Maximize z= 10000x+5000y----------(1)

             x≥100 ( since a shoe store owner wants to fill his shop with at least    100 pairs of men's shoes )

            y≥150  (at least 150 pairs of women's shoes).

also x+y≤400   (The store can only accommodate 400 pairs of shoes).

         x≤150    (male shoes should not exceed 150 pairs).

⇒         100≤x≤150

Hence the optimal solution of an LPP always lie on the end points.

We have end points as:

(100,300), (150,250), (100,150),(150,150).

by putting these value in equation (1) we see which give the maximum solution.

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for (100,150) i.e. x=100 and y=150: z=1,750,000

for (150,150) i.e. x=150 and y=150: z=2,250,000

Hence maximum profit is obtained at x=100 and y=300.

i.e. to maximize the profit:

<em>Number of men's shoes to be kept in store=100</em>

<em>and number of women's shoes to be kept in store=300</em>

<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.

5 0
3 years ago
Please show me the work thank you
KiRa [710]
Answer: B. x < 25

Remember: when multiplying or dividing negatives, flip the inequality sign.

3(x + 3) > 4(x - 4)                 Distributive Property
3x + 9 > 4x - 16                   Subtract 4x from both sides
-x + 9 > -16                          Subtract 9 from both sides
-x > -25                               Divide by -1 to both sides and flip the sign
x < 25                                 Answer!
 
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