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Rama09 [41]
1 year ago
6

I have to take a pic of the question

Mathematics
1 answer:
11111nata11111 [884]1 year ago
8 0

ANSWER

y=\frac{2}{3}x+\frac{4}{3}

EXPLANATION

We want to find the equation of the straight line given on the graph.

The general form of a linear equation is given as:

y=mx+b

where m = slope; b = y-intercept

To find the slope, we apply the formula for slope:

m=\frac{y_2-y_1}{x_2-x_1}

where (x1, y1) and (x2, y2) are two points on the line

Let us pick (1,2) and (4,4)

Therefore, the slope is:

m=\frac{4-2}{4-1}=\frac{2}{3}

To find the equation, we now apply the point-slope formula:

y-y_1=m(x-x_1)

Therefore, we have that the equation of the line is:

\begin{gathered} y-2=\frac{2}{3}(x-1) \\ y-2=\frac{2}{3}x-\frac{2}{3} \\ y=\frac{2}{3}x-\frac{2}{3}+2 \\ y=\frac{2}{3}x+\frac{4}{3} \end{gathered}

That is the equation of the line.

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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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