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3 years ago

Linear equations. write the standard form of an equation of the

Mathematics

1 answer:

erastova [34]3 years ago

7 0

Answer:

The standard form of an equation of the line that passes through the given point having slope is $y-y_{1} =m(x-x_{1}$

Step-by-step explanation:

The standard form of an equation of the line that passes through the given point ( $(x_{1} ,y_{1} )$ and having slope m is

$y-y_{1} =m(x-x_{1}$

another form of an equation of the line that passes through the given point ( $(x_{0} ,y_{0} )$ and having slope m is

$y-0 =m(x-0)$

y=m x

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Citrus2011 [14]

Ans. 1 Angle E and angle F are corresponding angles.

Solution 2.

So,

angle E = angle F

=> x + 20 = 5x - 20

=> x + 20 + 20 = 5x

=> x + 40 = 5x

=> 40 = 5x - x

=> 40 = 4x

=> 40/4 = x

=> 10 = x

Solution 3.

E = x + 20

=> E = 10 + 20

=> E = 30

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=> F = 5(10) - 20

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=> F = 30

Again justified that they are equal ( besides answer 1).

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3 years ago

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RUDIKE [14]

The perimeter, by definition, is the outside measure of that figure. MN and LM are the same length and LK and NK are the same length....we just need to find the lengths! Use the distance formula to find the distance between the 2 points:
$\sqrt{( x_{2} - x_{1} ) ^{2} +( y_{2}- y_{1} ) ^{2} }$
For the segment MN, use the coordinates of M as your x1, y1, and use the coordinates of N for x2, y2:
$\sqrt{(3-2) ^{2}+(4-3) ^{2} }$
which simplifies to
$\sqrt{(1 )^{2}+(1) ^{2} }$ which is $\sqrt{2}$
So that is the length of both MN and LM. So far our perimeter is $\sqrt{2} + \sqrt{2}=2 \sqrt{2}$
Now let's use the same formula to find out the length of one of the longer segments:
$\sqrt{(5-3) ^{2} +(3-2) ^{2} }$
which simplifies down to
$\sqrt{(2) ^{2} +(1) ^{2} }$
which is of course $\sqrt{5}$
Since we have 2 of those lengths, $\sqrt{5} + \sqrt{5}=2 \sqrt{5}$
So our perimeter is, in the end, $2 \sqrt{2}+2 \sqrt{5}$
That's the third choice down

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