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2 years ago

What is the value of y−5x−3for x = 2 and y = -4

Mathematics

2 answers:

arlik [135]2 years ago

8 0

Answer

-17

<u>Explanation</u>

y−5x−3 for x = 2 and y = -4

We solve this question by substituting the values of y and x.

y−5x−3 = -4 - 5(2) - 3

= -4 - 10 -3

= -14 - 3

= -17

svet-max [94.6K]2 years ago

4 0

y−5x−3 for x = 2 and y = -4

To find the value of y−5x−3 , we plug in 2 for x and -4 for y

y−5x−3

-4 - 5(2) - 3

Now we use order of operation and simplify it

We remove the parenthesis first

-4 - 10 -3

Now we add all the numbers and put the same sign because all the numbers are negative

-17

The value of y−5x−3 is -17

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The volume of a pyramid varies jointly as the height and the area of the base. If a pyramid has the measurements V=1144

Naddika [18.5K]

Answer:

11040 m³
k ≈ 0.33
V = (1/3)Bh

Step-by-step explanation:

The given relation is ...

V = kBh . . . . . for some base area B, height h, and constant of variation k

We are given length and width of the base so we presume it is a rectangle.

B = l·w = 8·11 = 88 . . . . square meters

The given volume tells us the value of k:

1144 = k(88)(39) . . . . . . cubic meters

1144/3432 = k = 1/3 ≈ 0.33

The value of k is about 0.33.

Then the volume of the larger pyramid is ...

V = (1/3)(15 m)(46 m)(48 m) = 11,040 m³

The general relationship is ...

V = 1/3Bh

7 0

2 years ago

Determine the equation of the line that passes through the point (7, -23)and is perpendicular to the line y=1/3x-6Enter your ans

marishachu [46]

We have a line y = 1/3x -6

We want a line that is perpendicular to this line

Perpendicular lines have slopes that multiply to -1

1/3 * m = -1

3 * 1/3 *m = -1 * 3

m = -3

The slope of the perpendicular line is -3

y = mx+b where m is the slope and b is the y intercept

y = -3x+b

We have a point on the line ( 7 ,-23)

Substitute this point into the equation

-23 = -3(7)+b

-23 = -21+b

Add 21 to each side

-23+21 = -21+21+b

-2 = b

y = -3x-2

In slope intercept form, the line perpendicular passing through (7,-23) is

y = -3x-2

3 0

7 months ago

2.5 ( y + ⅖ ) = -13<br> im really bad at math someone pls help

My name is Ann [436]

Answer:

-5.6

Step-by-step explanation:

first distribute the 2.5

also convert the 2/5 to decimal (0.4) because it's hard to work with both decimal and fraction

2.5y + 2.5*0.4 = -13

2.5y +1 = -13

y = -14/2.5 = -5.6

5 0

2 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

2 years ago

GEOMETRY PROBLEM THAT IS SOO HARD FOR ME <br> PLEASE HELP ME!!!!

12345 [234]

59.5 feet
Work:
17/10 = x/35
10x=595
59.5

3 0

2 years ago