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Nookie1986 [14]
3 years ago
5

A trust fund worth $25,000 is invested in two different portfolios, this year, one portfolio is expected to earn 5.25% interest

and the other is expected to earn 4%. Plans are for the total interest on the fund to be $1,150 in one year. How much money should be invested at each rate
Mathematics
2 answers:
Daniel [21]3 years ago
8 0

Answer:

Step-by-step explanation:

Let x be the investment at 5.25% and y at 4%

Then we have

x+y =25000\\5.25x+4y = 115000

We have to solve this system to get x and y

Multiply I equation by 4

100000=4x+4y\\115000=5.25x+4y\\-----------------------------\\15000 = 1.25x\\x = 12000\\y =25000-12000 = 13000

12000 should be invested in 5.25% and 13000 in 4% interest.

OLEGan [10]3 years ago
6 0

Answer:

The amount of money that should be invested at the rate of 5.25% is $12,000 and the amount money that should be invested at the rate of 4% is $13,000

Step-by-step explanation:

we know that

The simple interest formula is equal to

I=P(rt)

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

Let

x ------> the amount of money that should be invested at the rate of 5.25%

25,000-x -----> the amount money that should be invested at the rate of 4%

in this problem we have

t=1\ year\\ P_1=\$x\\P_2=\$(25,000-x)\\r_1=0.0525\\r_2=0.04\\I=\$1,150

substitute in the formula above

I=P_1(r_1t)+P_2(r_2t)

1,150=x(0.0525*1)+(25,000-x)(0.04*1)

Solve for x

1,150=0.0525x+1,000-0.04x

0.0525x-0.04x=1,150-1,000

0.0125x=150

x=\$12,000

|$25,000-x=\$13,000

therefore

The amount of money that should be invested at the rate of 5.25% is $12,000 and the amount money that should be invested at the rate of 4% is $13,000

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a. There are 0 or 2 real positive roots for the equation and

b. There are 0 or 2 real negative roots for the equation.

<h3>What is the Descartes'rule of sign?</h3>

Descartes' rule of sign states that

  • The number of real positive zero of a polynomial f(x) is the number of sign changes of the coefficients of f(x) or an even number less than the number of sign changes of the coefficients of f(x)
  • The number of real negative zero of a polynomial f(x) is the number of sign changes of the coefficients of f(-x) or an even number less than the number of sign changes of the coefficients of f(-x)

<h3>How to find the number of possible positive and negative roots are there for the equation?</h3>

Given the equation 0 = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1, writing it as a polynomial function, we have f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1

<h3>a. The number of positive roots</h3>

So, to find the number of positive roots, we find the number of sign changes of the polynomial f(x).

So, f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1

Since f(x) has coefficients -8, -2, + 8, -4, -1, there are two sign changes from -2 to + 8 and from + 8 to -4.

So, there are 2 or 2 - 2 = 0 real positive roots.

So, there are 0 or 2 real positive roots for the equation.

<h3>b. The number of negative roots</h3>

So, to find the number of negative roots, we find the number of sign changes of the polynomial f(-x).

So, f(-x) = −8(-x)¹⁰ − 2(-x)⁷ + 8(-x)⁴ − 4(-x)² − 1

= −8x¹⁰ + 2x⁷ + 8x⁴ − 4x² − 1

Since f(x) has coefficients -8, +2, + 8, -4, -1, there are two sign changes from -8 to + 2 and from + 8 to -4.

So, there are 2 or 2 - 2 = 0 real negative roots.

So, there are 0 or 2 real negative roots for the equation.

So,

  • There are 0 or 2 real positive roots for the equation and
  • There are 0 or 2 real negative roots for the equation.

Learn more about Descartes' rule of sign here:

brainly.com/question/28487633

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