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3 years ago

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Rick is thinking of a positive factor of $14$ and Steve is thinking of a positive factor of $42$. If Rick and Steve are thinking

of the same number, how many possible numbers could they be thinking of?

1 answer:

Black_prince [1.1K]3 years ago

4 0

Answer:

Step-by-step explanation:

Hello,

14 = 7 * 2 * 1

42 = 7 * 3 * 2 * 1

It can be 14, 7, 2 or 1

So there are 4 different positive numbers which meet the criteria

Hope this helps

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3 years ago

Read 2 more answers

The senior classes at High School A and High School B planned separate trips to Lake Compounce. The senior class at High School

storchak [24]

Answer:

Number of students in vans =15 and bus =46

Step-by-step explanation:

use variables for the vans and buses

let x= vans and y=buses

Now solve the 2 equations simultaneously

<em><u>remember</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>your</u></em><em><u> </u></em><em><u>equations</u></em>

6x+7y=412..........1

6x+9y=504.........2

make x the subject of one equation

and label it equation 3

6x=412—7y

$x = \frac{206}{3} - \frac{7y}{6} .............3$

subs 3 into 2

$6( \frac{206}{3} - \frac{7y}{6} ) + 9y = 504$

$412 - 7y + 9y = 504$

$- 7y + 9y = 504 - 412$

$2y = 92$

$y = 46...................4$

subs 4 into 1

$6x + 7(46) = 412$

$6x + 322 = 412$

$6x = 412 - 322$

$6x = 90$

$x = 15$

Therefore there are 15 students in each van and 46 students in each bus.

3 0

3 years ago

Which of the following sets of lengths can Sean use to make a right triangle?

Aleks [24]

Answer:

{8 cm, 15 cm, 17 cm}

Step-by-step explanation:

we know that

The length sides of a right triangle must satisfy the Pythagoras Theorem

so

$c^2=a^2+b^2$

where

c is the greater side (the hypotenuse)

a and b are the legs (perpendicular sides)

<u><em>Verify each case</em></u>

case 1) we have

{5 cm, 15 cm, 18 cm}

substitute in the formula

$18^2=5^2+15^2$

$324=250$ ----> is not true

therefore

Sean cannot make a right triangle with this set of lengths

case 2) we have

{6 cm, 12 cm, 16 cm}

substitute in the formula

$16^2=8^2+12^2$

$256=208$ ----> is not true

therefore

Sean cannot make a right triangle with this set of lengths

case 3) we have

{5 cm, 13 cm, 15 cm}

substitute in the formula

$15^2=5^2+13^2$

$225=194$ ----> is not true

therefore

Sean cannot make a right triangle with this set of lengths

case 4) we have

{8 cm, 15 cm, 17 cm}

substitute in the formula

$17^2=8^2+15^2$

$289=289$ ----> is true

therefore

Sean can make a right triangle with this set of lengths

6 0

3 years ago