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Sladkaya [172]
3 years ago
10

Please help me with my homework! Its past due and I am having trouble on it :/

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
5 0

Answer:

the first one is 57

the second one is 60

3rd one is 13

4th one is 9

Step-by-step explanation:

hope this helps :) have a nice day !!

**please let me know if this was wrong**

Nezavi [6.7K]3 years ago
4 0

Answer:

1) In the second step, it was supposed to multiply 9 and 2 and 36 by 3 instead of adding 9 and 3 to each other because multiplication and division comes before adding, according to pemdas.  

2) In the second step, they were supposed to multiply 6 and 4 instead of subtracting 27 and 6 because multiplication comes before subtraction, according to pemdas.

3)

(43 - 20) - 2 · 5

43 - 20 = 23

23 - 2 · 5

2 · 5 = 10

23 - 10 =

13

4) 81 ÷ (13 - 4)

13 - 4 = 9

81 ÷ 9

81 ÷ 9 =

9

5) 7 · 8 - 3 · 6 + 1

7 · 8 = 56

3 · 6 = 18

56 - 18 + 1

56 - 18 = 38

38 + 1 =

39

6)

7)

8)

Step-by-step explanation:

<h3>Section 1: Explain the error in each problem below. </h3>

1) Pemdas says multiplication and division comes before adding, so you must solve the multiplication in the problem first before the addition.  

2) Again, as pemdas says, multiplication comes first before subtraction, so you must solve the multiplication in the problem first before the sutraction.

<h3>Section 2: Simplify the following expressions. </h3>

3) (43 - 20) - 2 · 5

Pemdas says to solve the parenthesis first.

  • 43 - 20 = 23
  • 23 - 2 · 5

Now multiplication comes next (because there's no exponents).

  • 2 · 5 = 10
  • 23 - 10

Now subtraction.

  • 23 - 10 = 13

4) 81 ÷ (13 - 4)

Pemdas says to solve the parenthesis first.

  • 13 - 4 = 9
  • 81 ÷ 9

Now division comes next.

  • 81 ÷ 9 = 9

5) 7 · 8 - 3 · 6 + 1

Pemdas says to solve the multiplication first.

  • 7 · 8 = 56
  • 3 · 6 = 18
  • 56 - 18 + 1

Now subtraction and addition.

  • 56 - 18 = 38
  • 38 + 1 = 39

6)

<h3>Section 3: Insert grouping symbols to make the statements true.  </h3>

7)

8)

P.S. I am updating

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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

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Therefore, the length between points AB is

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

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Similarly, length BC is given by

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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)|

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C(2,2)

D'(3, -1)

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Area of polygon ABCD' = 19.5 units².

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