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3 years ago

Which number is a common multiple of 7and 4 What is the least common multiple of 18and28

Mathematics

1 answer:

lianna [129]3 years ago

8 0

The least common multiple of 18 and 28 is 252
the common multiple of 7 and 4 is 28

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What is the answer to 2/3 (5x – 5) = 10 because i don't know it

mars1129 [50]

Answer:

x = 76/75

Step-by-step explanation:

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3 years ago

⦁ Evaluate ⦁ 6! ⦁ 8P5 ⦁ 12C4

lisabon 2012 [21]

6! = 6 * 5 *4*3*2*1 =720

8p5 = 8*7*6*5*4 = 6720

12C4 = (12p4)/4! = (12*11*10*9)/4*3*2*1 = 495

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3 years ago

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Margarita [4]

Answer :-

9(3+√3) feet

Step by step explanation :-

A triangle is given to us. In which one angle is 30° and length of one side is 18ft ( hypontenuse) .So here we can use trignometric Ratios to find values of rest sides. Let's lable the figure as ∆ABC .

Now here the other angle will be = (90°-30°)=60° .

In ∆ABC ,

=> sin 30 ° = AB / AC

=> 1/2 = AB / 18ft

=> AB = 18ft/2

=> AB = 9ft .

Again In ∆ ABC ,

=> cos 30° = BC / AC

=> √3/2 = BC / 18ft

=> BC = 18 * √3/2 ft

=> BC = 9√3 ft .

Hence the perimeter will be equal to the sum of all sides = ( 18 + 9 + 9√3 ) ft = 27 + 9√3 ft = 9(3+√3) ft .

<h3>Hence the perimeter of the triangular pathway shown is 9 ( 3 + √3 ) ft .</h3>

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3 years ago

Arthur makes $216 each week writing a column for a newspaper. He recently wrote his first book and earns $7 for each copy sold.

yuradex [85]

216+7c=517; 43 copies .... of he made 216 a week from the newspaper and and made 7 dollars per copy sold of his book it should be 216+7c=517 then you subtract 216 from both sides of the equation so 7c=301 then divide 301 by 7 = 43

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3 years ago

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To pass a test 36% marks are needed out of maximum a student gets 123 marks and is declared failed by 39 marks what are maximum

IgorLugansk [536]

If the student got 123 marks and failed by 39, your first step would be to add the two together, getting 162. This is the amount of marks needed to pass the test.

You would then divide 162 by .36 (.36 is equivalent to 36%). This would result in a maximum of 450 marks.

The maximum marks on this test are 450 marks.

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3 years ago