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Masteriza [31]
2 years ago
7

Sarah went to the movies.After the movies , she wanted to ride the bus home.The b6 and the b9 buses both stop at the movies outs

ide Sarahs houses
Mathematics
1 answer:
erastova [34]2 years ago
6 0

The next time the buses will leave together is at 3:36 p.m.

<h3>What is the least common multiple?</h3>

A least common multiple is the smallest number divisible by two or more numbers. This concept is important for this question because the least common multiple shows how often do both buses leave at the same time.

<h3>How to find the least common multiple?</h3>

1. List the multiples of each number.

  • B 9 = 12 minutes                           B6= 9 minutes
  • 12 x 1 = 12                                      9 x 1 = 9
  • 12 x 2 = 24                                    9 x 2 = 18
  • 12 x 3 = 36                                    9 x 3 = 27
  • 12 x 4 = 48                                    9 x4 = 36

2. Identify the common number.

If you check carefully, the common number is 36, this means every 36 minutes the two buses leave at the same time.

<h3>What is the next time both of the buses will leave together?</h3>

The next time will be at 3:36 p.m. Let's check this answer.

B9

  • 3: 00 p.m.
  • 3: 12 p.m.
  • 3: 24 p.m.
  • 3: 36 p.m.

B6

  • 3: 00 p.m.
  • 3: 09 p.m.
  • 3: 18 p.m.
  • 3: 27 p.m.
  • 3: 36 p.m.

Note: This question is incomplete because some of the information about the buses is not given; here is the missing section.

-A B9 and B6 bus both left the movies at 3 pm.

-B9 buses are scheduled to leave the movies every 12 minutes.

- B6 buses are scheduled to leave the movies every 9 minutes.

If the buses ran on schedule, what is the next time both of the buses will leave together?

Learn more about least common multiple in: brainly.com/question/13696879

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Solve for x:a(a²+b²)x²+b²x-a​
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Step-by-step explanation:

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Use the quadratic equation formula:

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}

1. Evaluate the discriminant D

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2. Solve for x

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}

\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}

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