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Tanzania [10]
2 years ago
6

4/y+2=10/5y solve for y

Mathematics
1 answer:
pantera1 [17]2 years ago
8 0

Answer: y=−1 or y=2

Step-by-step explanation:

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Harold has 4 classes each morning. Each class is 1 hour long, and there are 10 minutes between classes. The first class is at 8
Viktor [21]

The fourth class ends at 12:30 pm

<em><u>Solution:</u></em>

Given that Harold has 4 classes each morning

Each class is 1 hour long, and there are 10 minutes between classes

The first class is at 8 A.M

<em><u>To find: Time at which fourth class ends</u></em>

Since each is 1 hour long and 10 minutes gap between classes

First class = 8 A.M to 9 A.M

Second class = 9:10 A.M to 10 : 10 AM

Third class = 10 : 20 AM to 11 : 20 AM

Fourth class = 11 : 30 AM to 12 : 30 PM

Thus the fourth class ends at 12:30 pm

4 0
2 years ago
14-5+20÷2^2+4 wich order
Sladkaya [172]

14-5+20÷2∧2+4

exponent first: 14-5+20÷4+4

Divide: 14-5+5+4

Solve left to right: 9+5+4

add all: 14+4 = 18

8 0
3 years ago
What are 91.6%,0.91,11/12,0.917,9.2% least to greatest
erik [133]

Answer:

9.2, 11/12, 91.7, 91.6, .91

Step-by-step explanation:

7 0
3 years ago
Which series of transformations will not map figure H onto itself
7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation (x+0,y-2) will map vertices of the square into points

  • (2,1)\rightarrow (2,-1);
  • (1,2)\rightarrow (1,0);
  • (2,3)\rightarrow (2,1);
  • (3,2)\rightarrow (3,0).

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

  • (2,-1)\rightarrow (2,3);
  • (1,0)\rightarrow (1,2);
  • (2,1)\rightarrow (2,1);
  • (3,0)\rightarrow (3,2)

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation (x+2,y-0) will map vertices of the square into points

  • (2,1)\rightarrow (4,1);
  • (1,2)\rightarrow (3,2);
  • (2,3)\rightarrow (4,3);
  • (3,2)\rightarrow (5,2).

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

  • (4,1)\rightarrow (2,1);
  • (3,2)\rightarrow (3,2);
  • (4,3)\rightarrow (2,3);
  • (5,2)\rightarrow (1,2)

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation (x+3,y+3) will map vertices of the square into points

  • (2,1)\rightarrow (5,4);
  • (1,2)\rightarrow (4,5);
  • (2,3)\rightarrow (5,6);
  • (3,2)\rightarrow (6,5).

2nd transformation = reflection over y = -x + 7 will map vertices into points

  • (5,4)\rightarrow (3,2);
  • (4,5)\rightarrow (2,3);
  • (5,6)\rightarrow (1,2);
  • (6,5)\rightarrow (2,1)

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation (x-3,y-3) will map vertices of the square into points

  • (2,1)\rightarrow (-1,-2);
  • (1,2)\rightarrow (-2,-1);
  • (2,3)\rightarrow (-1,0);
  • (3,2)\rightarrow (0,-1).

2nd transformation = reflection over y = -x + 2 will map vertices into points

  • (-1,-2)\rightarrow (4,3);
  • (-2,-1)\rightarrow (3,4);
  • (-1,0)\rightarrow (2,3);
  • (0,-1)\rightarrow (3,2)

These points are not the vertices of the initial square.

5 0
3 years ago
What type a problem can be. Using the greatest common factor
sergeinik [125]
The word greatest in a word problem tells you to do GCF and the word least in a problem means to use the LCM
6 0
3 years ago
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