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3 years ago

4x - 10 > 22 or x-5 < -6 I need to know how to do this step by step..

Mathematics

2 answers:

diamong [38]3 years ago

6 0

Answer:

x > 3 or x < -1

Step-by-step explanation:

4x - 10 > 22

-10 -10

4x > 12

divide both sides by 4

x > 3

x - 5 < -6

+5 +5

x < -1

Hope this helps!!!

Sunny_sXe [5.5K]3 years ago

5 0

Ok so the answer is x>8 the steps would be
Calculate 4x> 22+ 10
The divide from both sides 4x>32 then the answer would be x>8

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Solve for the given time.

AlekseyPX

9514 1404 393

Answer:

$319,525.64

Step-by-step explanation:

It appears you want the interest to accumulate for 15 years before you pay anything on the loan. The value of $180,000 appreciating at 3.9% annually will be ...

$180,000·(1 +0.039)^15 = $319,525.64

If your lender allows you to wait 15 years before making any payment, your payment will be $319,525.64.

Additional comment

Home loans are usually paid with monthly payments computed from the formula ...

A = P(r/12)/(1 -(1 +r/12)^(-12t))

where A is the monthly payment on principal P, r is the annual interest rate, t is the number of years.

For your $180,000 loan, the monthly payment is ...

A = $180,000·(0.039/12)/(1 -(1 +0.039/12)^(-12·15)) ≈ $1322.44

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If you pay $1322.44 monthly on the loan, your total payment is $238,039.20.

3 0

3 years ago

The area of a rectangle is 40 square feet. What could be the perimeter of the rectangle? how is this problem solved.

schepotkina [342]

A possible perimeter for the rectangle would be 28

Area=L x W

40/10=4

Width=4, Length=10

Perimeter=L+L+W+W

4+4+10+10=28

4 0

3 years ago

Read 2 more answers

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: