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3 years ago

Please help

Mathematics

1 answer:

Firdavs [7]3 years ago

3 0

Answer:

(–10, –2), (6, 6)

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Verify each case

case 1) (5, –4), (–2, 1)

substitute in the formula

$m=\frac{1-(-4)}{-2-5}$

$m=\frac{1+4}{-2-5}$

$m=\frac{5}{-7}$

$m=-\frac{5}{7}$ ----> the slope is negative

case 2) (6, –10), (2, 10)

substitute in the formula

$m=\frac{10-(-10)}{2-6}$

$m=\frac{10+10}{2-6}$

$m=\frac{20}{-4}$

$m=-5$ ----> the slope is negative

case 3) (–10, –2), (6, 6)

substitute in the formula

$m=\frac{6-(-2)}{6-(-10)}$

$m=\frac{6+2}{6+10}$

$m=\frac{8}{16}$

$m=0.5$ ----> the slope is positive

case 4) (5, –1), (–6, 6)

substitute in the formula

$m=\frac{6-(-1)}{-6-5}$

$m=\frac{6+1}{-6-5}$

$m=\frac{7}{-11}$

$m=-\frac{7}{11}$ ----> the slope is negative

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OLga [1]

Answer:

x=4.

Step-by-step explanation:

Let's solve your equation step-by-step.

5/15=x/ x+8

Step 1: Cross-multiply.

5/15=x/ x+8

5*(x+8)=x*(15)

5x+40=15x

Step 2: Subtract 15x from both sides.

5x+40−15x=15x−15x

−10x+40=0

Step 3: Subtract 40 from both sides.

−10x+40−40=0−40

−10x=−40

Step 4: Divide both sides by -10.

−10x−10=−40−10

Then your answer will be x=4.

6 0

3 years ago

the cost of plastering a wall of a room having length 8m and heights 4m at Rs 15 per m² is Rs 1560 . find the cost of carpeting

Effectus [21]

Step-by-step explanation:

here ,

length = 8m

height=4m

Rate of plastering 4 walls = 15/m^2

Cost of plastering 4 wall = Rs 1560

Area of floor = Cost of plastering / Rate of plastering

= 1560/15

= 104

again,

Area of four wall = 2h(l+b)

or, 104= 2×4(8+b)

or, 104 =8(8+b)

or, 104/8=8+ b

or,13 - 8= b

or, b= 5

again,

Area of floor= l×b

= 8×5

= 40

again,

Rate of carpeting the floor= 300/ m^2

Area of floor = 40 m^2

cost of carpeting=Rate × Area

= 300×40

= 12000

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3 years ago

The equation of the line of best fit of a scatter plot is y = 6x − 9. What is the slope of the equation? –6 –9 9 6

Vinil7 [7]

6 is the answer because it is 6x therefore the gradient increases by 6 each time

5 0

3 years ago

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the 25 members of the reading club are buying a book to discuss at the next meeting but Club leader got a special price the book

Sever21 [200]

25*8.14=?

Multiply the cost of the book per person ($8.14) by the number of people (25).

6 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:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

3 years ago