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alexgriva [62]
3 years ago
11

Can someone please help me I really need help help me thank you

Mathematics
1 answer:
harkovskaia [24]3 years ago
5 0

Answer:

<h2><em><u>A to C = 25 </u></em></h2><h2><em><u>A to B = 13 </u></em></h2><h2><em><u>C to B = 37 </u></em></h2><h2><em><u></u></em></h2>

Step-by-Step Explanation:

<em><u>Perimeter</u></em> = 75

<em><u>Sides:</u></em>

2x + 3

3x + 4

2x - 9

<h2 /><h2><em><u>1. Equal the sides added together to the perimeter</u></em></h2>

75 = 2x + 3 + 3x + 4 + 2x - 9

<h2><em><u>2. Simplify Like terms</u></em></h2>

2x + 3 + 3x + 4 + 2x - 9  = 7x - 2

<h2><em><u>3. Place the equation back together</u></em></h2>

75 = 7x - 2

<h2><em><u>4. Isolate the variables and numbers</u></em></h2>

75 = 7x - 2

+2        +2

77 = 7x

<h2><em><u>5. Simplify the equation</u></em></h2>

77 = 7x

/7     /7

<h2><em><u>11 = x </u></em></h2>

<h2><em><u>6. Substitute the value of x into the side lengths.</u></em></h2>

2x + 3 = 2(11) + 3 = 22 + 3 = <em><u>25</u></em>

3x + 4 = 3(11) + 4 = 33 + 4 = <em><u>37</u></em>

2x - 9 = 2(11) - 9 = 22 - 9 = <em><u>13</u></em>

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PLEASE HELP WILL MARK BRAINLIEST!!(plz show work too)
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\huge \bf༆ Answer ༄

Let the capacity of bus be x students

And van be y students, now ;

From the given statements we get two equations ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \:   \:  \:  \: \:  (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:13x + 5y = 670 \:  \:  \:  \:  \: (2)

multiply the equation (2) with 2 [ it won't change the values ]

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \: \:  \:  \:   \:  \:  \:  \:  \: (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y = 1340 \:  \:  \:  \:  \: (3)

Now, deduct equation (1) from equation (3)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y - 2x - 10y = 1340 - 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:24x = 1080

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 1080 \div 24

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 45

Therefore each bus can carry (x) = 45 students

Now, plug the value of x in equation (1) to find y ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(2 \times 45) + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:90 + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 260 - 90

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 170

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 170 \div 10

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 17

Hence, each van can carry (y) = 17 students in total.

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