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boyakko [2]
3 years ago
7

Jose uses the pattern in rows 1 and 8 in the multiplication table below to find ratios that are equivalent to 1:8

Mathematics
2 answers:
kondor19780726 [428]3 years ago
8 0

Answer:

Step-by-step explanation:

its 80 i did the test

vladimir2022 [97]3 years ago
6 0

Answer:

it is 80

Step-by-step explanation:

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Find the area of a regular octagon with an apothem of 7 inches and a side length of 5.8 inches. (nearest tenth)
Kisachek [45]

Answer:

162.4 in²

Step-by-step explanation:

LETS GET INTOOOOEEETTT

Let's start with what we know:

Area of regular octagon = 1/2 x perimeter x apothem

We know the apothem, so all that we need to find to fill in the above equation is the perimeter:

perimeter = 8 x 5.8 = 46.4in

Now we can fill in our original equation and solve:

Area of regular octagon = 1/2 x perimeter x apothem

Formula = n (s/2)² divided by tan( π /n)

   = 8 (5.8/2)²  divided by tan ( π /8)

                 = 162.4283 in²

ORRR when rounded to the nearest tenth,

                      =162.4 in²

6 0
3 years ago
Sketching a Hyperbola In Exercises 19-32, find the center, vertices, foci, and the equations of the asymptotes of the hyperbola.
zlopas [31]

Step-by-step explanation:

her this is the answer for the question

7 0
2 years ago
True-False: Please select true or false and click "submit."
harina [27]
A true


because 53•(41•11)=23,903 and (53•41)•11=23,903


Hope this helped
4 0
3 years ago
Read 2 more answers
3 1/7 x 10 answer plz
Yanka [14]

Answer:

31.43

Step-by-step explanation:

3 1/7 = 3.14285714285

3 1/7 x 10 = 31.4285714286 which is about 31.43

4 0
2 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:

  • (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
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