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3 years ago

What is the total interest on a ten-year. 6.1% loan with a principal of $32,000?

1 answer:

6 0

We are given with:
Principal, P = $32000
Interest rate, i= 6.1%
Duration of loan, n= 10 years

To solve for the total interest, I, we use the formula:
I = P(1+i)^n - P
Substituting the given values:
I = 32000(1+.061)^10 - 32000
I = $25,850.06

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What is 80 in its simplest form as a fraction?

Ann [662]

I believe the correct given problem is:

“What is 80% in its simplest form as a fraction?”

Take note that you missed to place the percent sign which is very important.

Now to solve this, 80% when converted to decimal form is 0.80 which means that you have 80 parts per 100 parts. Therefore the fraction form of this would be 80 / 100.

fraction = 80 / 100

However this fraction is not yet the answer because we can still simplify this further. First step in simplifying is to divide both the numerator and denominator by 10 so that we are left with:

8 / 10

Next step is to divide by 2:

4 / 5

Now we can see that we can no longer divide this (numbers should be whole number). Therefore the simplest form as a fraction is 4/5

Answer:

4/5

6 0

3 years ago

Given f(x)=-3x+3solve for x when f(x)=6

natali 33 [55]

Answer:

x = -1

Step-by-step explanation:

Given: $f(x) = - 3x + 3$

Solve for x, When: $f(x) = 6$

Step by step:

$- 3x + 3 = 6$

$- 3x = 6 - 3$

$- 3x = 3$

$x = 3 \div - 3$

$\boxed{\green{x = - 1}}$

3 0

3 years ago

1. perpendicular to x - 6y = 2; (2, 4)

nordsb [41]

Answer:

all work is pictured and shown

7 0

3 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares: