Please help on 10 thank you so michy

Emily would like to buy some new

bonufazy [111]

Answer:

$14,277.80

Step-by-step explanation:

The standard formula for compound interest is given as;

A = P(1+r/n)^(nt) .....1

Where;

A = final amount/value

P = initial amount/value (principal)

r = rate yearly

n = number of times compounded yearly.

t = time of investment in years

For this case;

P = $7,400

t = 8 years

n = 4 (quarterly)

r = 9.5% = 0.095

Using equation 1.

A = $7,400(1+0.095/4)^(4×7)

A = $7,400(1.02375)^(28)

A = $7,400(1.929432606035)

A = $14,277.80

final amount/value after 8 years A =$14,277.80

6 0

2 years ago

2(x-3)-8< -4 I need a explanation and the answer please.

tresset_1 [31]

Answer:

The solution to the inequality is:

$x$

The number line solution graph is also attached below.

Step-by-step explanation:

Given the inequality

$2(x-3)-8< -4$

Add 8 to both sides

$2\left(x-3\right)-8+8$

$2\left(x-3\right)$

Divide both sides by 2

$\frac{2\left(x-3\right)}{2}$

$x-3$

Add 3 to both sides

$x-3+3$

$x$

Thus, the solution to the inequality is:

$x$

The number line solution graph is also attached below.

6 0

2 years ago

What dose 5d-10= using two word expressions

Naddik [55]

Maybe 5 multiplying or adding the value of d subtracting 10 equals

8 0

2 years ago

Factor by grouping 9m3-3m2p2-3mp+p3

Setler [38]

(3m^2-p)(3m-p^2)
You factor out the 3m^2 and p from the combined like terms and then group those into a factor and group the two identical factors together and (3m^2-p)(3m-p^2) is your answer.

3 0

3 years ago

Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is

ohaa [14]

The vectors in $S$ form a basis of $P_2$ if they are mutually linearly independent and span $P_2$ .

To check for independence, we can compute the Wronskian determinant:

$\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0$

The determinant is non-zero, so the vectors are indeed independent.

To check if they span $P_2$ , you need to show that any vector in $P_2$ can be expressed as a linear combination of the vectors in $S$ . We can write an arbitrary vector in $P_2$ as

$p=ax^2+bx+c$

Then we need to show that there is always some choice of scalars $k_1,k_2,k_3$ such that

$k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p$

This is equivalent to solving

$(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c$

or the system (in matrix form)

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

This has a solution if the coefficient matrix on the left is invertible. It is, because

$\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0$

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}$

Then

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}$

$\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}$

A solution exists for any choice of $a,b,c$ , so the vectors in $S$ indeed span $P_2$ .

The vectors in $S$ are independent and span $P_2$ , so $S$ forms a basis of $P_2$ .

5 0

2 years ago