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3 years ago

3 18/15. which as a proper fration is

Mathematics

1 answer:

stepladder [879]3 years ago

4 0

Answer:

You can't change 3 18/15 to an proper fraction

srry :(

Step-by-step explanation:

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If the coordinate of A is (0,-2) and the coordinate of B is (10,-6), then the midpoint of AB is (______).

balu736 [363]

Answer:

5, -4

Step-by-step explanation:

midpoint = x1 + x2/2 ,y1 + y2/2

= 0 + 10 /2 , -2 + -6/2

= 10/2 , -8/2

midpoint = 5, -4

4 0

3 years ago

Zeller, Inc., has 50,000 shares of stock outstanding. Zeller declares a dividend of $480,000. What is the dividend per share?

Monica [59]

48÷5 =9.6 hope it helps :-)

4 0

3 years ago

Solve for me please thank you

hammer [34]

The answer to those options is of course C

5 0

3 years ago

EEF has vertices D(1, 3), E(2, -4), and F(-3, 1). What are the vertices T<4, 1> (DEF).

Triss [41]

The vertices T<4, 1> (DEF) are D' = (5, 4), E' = (6, -3) and F' = (1, 2)

<h3>How to determine the vertices?</h3>

The vertices of the triangle are given as:

D(1, 3), E(2, -4), and F(-3, 1).

The transformation represented by T<4, 1> (DEF) is:

(x,y) = (x + 4, y + 1)

So, we have;

D' = (1 + 4, 3 + 1)

D' = (5, 4)

E' = (2 + 4, -4 + 1)

E' = (6, -3)

F' = (-3 + 4, 1 + 1)

F' = (1, 2)

Hence, the vertices T<4, 1> (DEF) are D' = (5, 4), E' = (6, -3) and F' = (1, 2)

Read more about transformation at:

brainly.com/question/1548871

#SPJ1

8 0

2 years ago

The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo

shepuryov [24]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY = $\sqrt{x^2+y^2}$

Therefore, the length between points AB is

length AB = $\sqrt{(-4 - (-2))^2+(-1-3)^2}$

= $\sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2} = \sqrt{20}$ = 4.472 units

Similarly, length BC is given by

Length BC = $\sqrt{(-4)^2+1^2} = \sqrt{17}$ = 4.123 units

Length CD = $\sqrt{(-2)^2+5^2} = \sqrt{29}$ = 5.385 units

Length DA = $\sqrt{(8)^2+(-2)^2} = \sqrt{68}$ = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A = $\sqrt{(7)^2+(0)^2} = \sqrt{49}$ =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

$S_{\bigtriangleup}$ = $(\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|$

= $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

$S_{{\bigtriangleup}ABC}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|$

= $=(\frac{1}{2})|-4-6 -8|= 9 units^2$

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

$S_{{\bigtriangleup}ACD'}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2}$ = 10.5 units²

Area of polygon = $S_{{\bigtriangleup}ABCD'}$ = $S_{{\bigtriangleup}ABC}$ + $S_{{\bigtriangleup}ACD'}$ = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

4 0

3 years ago