All categories

2 years ago

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:

You might be interested in

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

Solution:

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

To find: Time at which fourth class ends

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:

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