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Inessa [10]
3 years ago
9

Rafael's art class is 6 hours long. How long is his class in minutes?

Mathematics
2 answers:
sdas [7]3 years ago
6 0

Answer:

The art class would be 360 mins long.

Step-by-step explanation:


spin [16.1K]3 years ago
3 0

<u>Answer</u>:

360 Minutes Long

<u>Step-by-step explanation:</u>

There are 60 minutes in an hour. Multiply 60*6, and you get 360 minutes.

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100 points!!! Show your work on paper
lyudmila [28]

Answer:

  • 10 cm, 15 cm

Step-by-step explanation:

Set ratios and solve for x

<u>Corresponding sides are AC with A'C' and CB with C'B'</u>

  • AC/A'C' = CB/C'B'
  • (x - 1)/2 = (x + 4)/3
  • 3(x - 1) = 2(x + 4)
  • 3x - 3 = 2x + 8
  • 3x - 2x = 8 + 3
  • x = 11

<u>Find the values of AC and CB:</u>

  • AC = x - 1 = 11 - 1 = 10 cm
  • CB = x + 4 = 11 + 4 = 15 cm
5 0
2 years ago
Read 2 more answers
3. Which expressions are equivalent to the given one?
Montano1993 [528]

Answer:

b, d

Step-by-step explanation:

20/-4 = -5

20 x (-1/4) = -5

4 0
1 year ago
Can u answer these for me with the work shown
babymother [125]

Answer:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

Step-by-step explanation:

Required

Simplify

Solving (1):

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}

Factorize the numerator and the denominator

\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}

Factor out x+2 at the numerator

\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}

Express x^2 - 9 as difference of two squares

\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}

\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}

Expand the denominator

\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}

Factorize

\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}

\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}

Cancel out same factors

\frac{(x+3)}{x}

Hence:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

Solving (2):

\frac{3x^2 - 5x - 2}{x^3 - 2x^2}

Expand the numerator and factorize the denominator

\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}

Factorize the numerator

\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}

Factor out x - 2

\frac{(3x + 1)(x - 2)}{x^2(x- 2)}

Cancel out x - 2

\frac{3x + 1}{x^2}

Hence:

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

Solving (3):

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}

Express x^2 - 9 as difference of two squares

\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}

Factorize all:

\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}

Cancel out x + 3 and 3 + x

\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}

\frac{3 - x}{x- 3} * \frac{5}{2x}

Express 3 - x as -(x - 3)

\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\

-1 * \frac{5}{2x}

-\frac{5}{2x}

Hence:

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

Solving (4):

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}

Expand x^2 - 6x + 9 and factorize 5x - 15

\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}

Factorize

\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}

\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}

Cancel out x - 3

\frac{(x -3)}{5} / \frac{5}{3-x}

Change / to *

\frac{(x -3)}{5} * \frac{3-x}{5}

Express 3 - x as -(x - 3)

\frac{(x -3)}{5} * \frac{-(x-3)}{5}

\frac{-(x-3)(x -3)}{5*5}

\frac{-(x-3)^2}{25}

Hence:

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

Solving (5):

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}

Factorize the numerator and expand the denominator

\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}

Factor out x - 1 at the numerator and factorize the denominator

\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}

x +1

Hence:

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

Solving (6):

\frac{9x^2 + 3x}{6x^2}

Factorize:

\frac{3x(3x + 1)}{3x(2x)}

Divide by 3x

\frac{3x + 1}{2x}

Hence:

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

Solving (7):

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}

Change / to *

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Expand

\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Factorize

\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

Cancel out x - 2 and x - 1

\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}

Cancel out x

\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}

\frac{12x^2}{4x}

3x

Hence:

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

8 0
2 years ago
Mr. Thornton's class visited five freshwater lakes to learn more about the crocodiles and alligators living in them to class cou
Yuliya22 [10]

Given table is given below

The relationship between the number of crocodiles and the number of alligators is not proportional across all lakes. Only two of the lakes have crocodiles and alligators in the same proportion. Which two lakes are they?

lake A and lake B

lake B and lake C

lake A and lake E

lake C and lake D

Answer : lake A and lake E

Lake ---> Crocodiles ---> Alligators ---> Ratio

A ---> 4 ----> 5

B ---> 21 ---> 35

C ---> 3 ---> 9

D ---> 6 ---> 16

E ---> 24 ---> 30

The ratio of A is 4: 5

B is 21: 35 = 3 : 5

C is 3 : 9 = 1 : 3

D is 6 : 16 = 3 : 8

E is 24: 30 = 4 : 5

Lake A and E has the same ratio 4: 5

7 0
3 years ago
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I marked them on coordinate plane
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3 years ago
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