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1 year ago

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Mathematics

1 answer:

snow_lady [41]1 year ago

5 0

Answer:

$2000 was invested at 5% and $5000 was invested at 8%.

Step-by-step explanation:

Assuming the interest is simple interest.

<u>Simple Interest Formula</u>

I = Prt

where:

I = interest earned.
P = principal invested.
r = interest rate (in decimal form).
t = time (in years).

Given:

Total P = $7000
P₁ = principal invested at 5%
P₂ = principal invested at 8%
Total interest = $500
r₁ = 5% = 0.05
r₂ = 8% = 0.08
t = 1 year

Create two equations from the given information:

$\textsf{Equation 1}: \quad \sf P_1+P_2=7000$

$\textsf{Equation 2}: \quad \sf P_1r_1t+P_2r_2t=I\implies 0.05P_1+0.08P_2=500$

Rewrite Equation 1 to make P₁ the subject:

$\implies \sf P_1=7000-P_2$

Substitute this into Equation 2 and solve for P₂:

$\implies \sf 0.05(7000-P_2)+0.08P_2=500$

$\implies \sf 350-0.05P_2+0.08P_2=500$

$\implies \sf 0.03P_2=150$

$\implies \sf P_2=\dfrac{150}{0.03}$

$\implies \sf P_2=5000$

Substitute the found value of P₂ into Equation 1 and solve for P₁:

$\implies \sf P_1+5000=7000$

$\implies \sf P_1=7000-5000$

$\implies \sf P_1 = 2000$

$2000 was invested at 5% and $5000 was invested at 8%.

Learn more about simple interest here:

brainly.com/question/27743947

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Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$