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3 years ago

A $6000 investment earns $690 simple interest in two years. What is the annual interest rate?

1 answer:

6 0

2 years simple interest is $690
1 year simple interest is $690/2 = $345

$6000 simple interest is $345
$1000 simple interest is $345/6 = $57.5
$100 simple interest is $57.5/10 = $5.75

Annual interest rate is 5.75%

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Solve for x and show work

jekas [21]

Answer:

x = 6

Step-by-step explanation:

Angle RPG = (Arc RQ + Arc SC)/2 Interior Secant Theorem

124 = (197 + 9x - 3)/2 Substitution

124 = (194 + 9x)/2 Alg

124 = 97 + 4.5x Alg

27 = 4.5x Alg

x = 6 Alg

Hope this helped! Ask me if you didn't understand something.

7 0

3 years ago

A motor cycle purchased for $9,000 today will be worth 6% less each year. For what can you expect to sell the motorcycle at the

Salsk061 [2.6K]

Answer:

$5,956.2

Step-by-step explanation:

First we need to find the amount after 5years using the formula

A =P(1+r)^n

A = 9000(1+0.06)^5

A = 9000(1.06)^5

A = 9000(1.3382)

A = 12,043.8

Interest after 5 years = 12,043.8 - 9000

Interest after 5 years = 3,043.8

Since it will be worth 6% less each year, the amount after 5years will be 9000 - 3,043.8 = $5,956.2

6 0

3 years ago

PLS HELP I REALLY NEED HELP

lbvjy [14]

Answer:

a) AB = 10 units
b) Midpoint is (2, 6)

===================

<h3 /><h3>Given</h3>

Points A( - 1, 10) and B(5, 2)

a) The length of AB
b) The midpoint of AB

<h3>Solution</h3>

a) Use the distance formula:

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates and calculate:

$d=\sqrt{(5-(-1))^2+(2-10)^2} =\sqrt{6^2+(-8)^2} =\sqrt{36+64} =\sqrt{100} =10$

The distance is AB = 10 units

b) Use midpoint formula and find x and y- coordinates of this point:

$x= \cfrac{x_1+x_2}{2}$ and $y= \cfrac{y_1+y_2}{2}$

Substitute coordinates and find the midpoint:

$x= \cfrac{-1+5}{2} =2$ and $y= \cfrac{10+2}{2}=6$

The midpoint is (2, 6)

3 0

2 years ago

Read 2 more answers

Factor the polynomial, x2 + 5x + 6

patriot [66]

Answer:

Choice b.

$x^{2} + 5\, x + 6 = (x + 3)\, (x + 2)$ .

Step-by-step explanation:

The highest power of the variable $x$ in this polynomial is $2$ . In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to $0$ .)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to $0$ . The discriminant of this polynomial is $(5^{2} - 4 \times 1 \times 6) = 1$ .

$\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}$ .

Similarly:

$\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}$ .

By the Factor Theorem, if $x = x_{0}$ is a root of a polynomial, then $(x - x_0)$ would be a factor of that polynomial. Note the minus sign between $x$ and $x_{0}$ .

The root $x = -2$ corresponds to the factor $(x - (-2))$ , which simplifies to $(x + 2)$ .
The root $x = -3$ corresponds to the factor $(x - (-3))$ , which simplifies to $(x + 3)$ .