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Lady_Fox [76]
3 years ago
15

Find the equation of the graphed line.

Mathematics
1 answer:
belka [17]3 years ago
4 0
<h3><u>Answer:</u></h3>

\boxed{\boxed{\pink{\bf \leadsto Hence \ option\ [d]\ \bigg(y =  \dfrac{5}{2}x + 5\bigg) \ is \ correct  }}}

<h3><u>Step-by-step explanation:</u></h3>

Here from the given graph we can see that the graph the graph intersects x axis at (2,0) and y axis at (5,0). On seeing options it's clear that we have to use Slope intercept form . Which is :-

\large\boxed{\orange{\bf y = mx + c} }

We know that slope is \tan\theta. So here slope will be ,

\red{\implies slope = \tan\theta} \\\\\implies slope =\dfrac{PERPENDICULAR}{BASE} \\\\\bf \implies slope =\dfrac{5}{2}

Hence the slope is 5/2 . And here value of c will be 5 since it cuts y axis at (5,0).

\purple{\implies y = mx + c }\\\\\implies y = \dfrac{5}{2}x + 5 \\\\\underline{\boxed{\red{\tt \implies y = \dfrac{5}{2}x + 5 }}}

<h3><u>Hence</u><u> </u><u>option</u><u> </u><u>[</u><u> </u><u>d</u><u> </u><u>]</u><u> </u><u>is</u><u> </u><u>corre</u><u>ct</u><u> </u><u>.</u><u> </u></h3>
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Rama09 [41]

You are solving for a stem-and-leaf plot that represents the given data: 80, 81, 91, 92, 66, 55, 54, 30, 55, 79, 78.

Note that a stem-and-leaf plot is used in this way:

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In this case, the stem is the tens place value, and is signified by 8. 7, on the other hand, is the ones place value. <em>Multiple leafs can share the same stem</em>. For example, a stem-and-leaf plot of 85, 86, 87, would give you 8|5 , 6 , 7.

Now, set the given data set in order:

30 , 54 , 55 ,  55 , 66 , 78 , 79 , 80 , 81 , 91 , 92

Now, create the stem-and-leaf plot:

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\left[\begin{array}{c|ccc}3&0\\5&4&5&5\\6&6\\7&8&9\\8&0&1\\9&1&2\end{array}\right]

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Your answer will be:

A stem value of 3, with a leaf value of 0. A stem value of 5, with a leaf value of 4 & 5. A stem value of 6, and a leaf value of 6. A stem value of 7, and a leaf value of 8 & 9. A stem value of 8, and a leaf value of 0 & 1. A stem value of 9, and a leaf value of 1 & 2.

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2 years ago
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3 years ago
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artcher [175]

Answer:

\boxed{y

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