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Delicious77 [7]
3 years ago
13

Without drawing the graph of the function y=1.2x−7, see if the graph passes through points: (100; 113)

Mathematics
1 answer:
madreJ [45]3 years ago
3 0

Answer:

(100,113) is on the graph of y = 1.2x -7

Step-by-step explanation:

y=1.2x−7

Substitute the point in and see if the equation is true

113 = 1.2 (100) -7

113 =120 - 7

113 = 113

This is true, so the point is on the graph

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Solve and check:8a-4=25+6(a-8)
OlgaM077 [116]
8a - 4 = 25 + 6 (a - 8)

(distribute the 6)

8a - 4 = 25 + 6a - 48

(combine like terms 25 - 48)

8a - 4 = 6a - 23

( subtract 6a from both sides)

2a - 4 = -23

(add 4 to both sides)

2a = -19

(divide both sides by 2)

a = - 9.5

////CHECK/////

8 (-9.5) - 4 = 25 + 6 ( -9.5 -8)

(plug each side into a calculator)

-80 = -80

FINAL ANSWER:

a = -9.5
5 0
2 years ago
Find the surface area of the right triangular prism (above) using its net (below).
lakkis [162]

Answer:

144 units²

Step-by-step explanation:

The net of the right triangular prism consists of 3 rectangles and 2 equal triangles

Let's solve for the area of each:

✔️Area of rectangle 1 = L*W

L = 11

W = 3

Area of rectangle 1 = 11*3 = 33 units²

✔️Area of rectangle 2 = L*W

L = 11

W = 4

Area of rectangle 2 = 11*4 = 44 units²

✔️Area of rectangle 3 = L*W

L = 11

W = 5

Area of rectangle 3 = 11*5 = 55 units²

✔️Area of the two triangles = 2(½*base*height)

base = 4

height = 3

Area of the two traingles = 2(½*4*3)

= 12 units²

✔️Surface area of the right triangle = area of rectangle 1 + area of rectangle 2 + area of rectangle 3 + area of the two triangles

= 33 + 44 + 55 + 12

= 144 units²

7 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
Use 3 for at Volume of a Cylinder
pychu [463]

Step-by-step explanation:

464.5cm³

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3 0
2 years ago
Given the system of equations, what is the y-coordinate of the solution?
bulgar [2K]
If you plug in the second equation (x=5-3/2y) into the first, you get:

5(5-3/2y) - 4y = 7

Simplify:

25 - 15/2y - 4y = 7
-15/2 y - 8/2 y = 7-25 
23/2 y = 18
y = 18 * 2/23 = 36/23

answer B!
6 0
2 years ago
Read 2 more answers
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