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Reil [10]
3 years ago
14

Write a linear equation from the ordered pair (-2,8)

Mathematics
1 answer:
OlgaM077 [116]3 years ago
4 0
The equation of a line that goes through the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)

so, given the point (-2,8)
x1=-2
y1=8

y-8=m(x-(-2))
y-8=m(x+2)
the equation is y-8=m(x+2) where m is the slope
or you could rewrite it to get y=mx+2m+8

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30 = X-8<br> One step equation
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Answer:

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Step-by-step explanation:

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3 years ago
The sum of two numbers is equal to 11 and the difference is 19. What are the two numbers?
charle [14.2K]

\bold{\huge{\pink{\underline{ Solution }}}}

<u>We </u><u>have </u><u>given </u><u>in </u><u>the </u><u>question </u><u>that</u><u>, </u>

  • <u>The </u><u>sum </u><u>of </u><u>2</u><u> </u><u>numbers </u><u>is </u><u>equal </u><u>to </u><u>1</u><u>1</u><u> </u>
  • <u>The </u><u>difference </u><u>between </u><u>two </u><u>numbers </u><u>is </u><u>1</u><u>9</u><u> </u><u>.</u>

\bold{\underline{ To \: Find }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y</u><u>. </u>

\bold{\underline{ Let's \: Begin }}

Let the two numbers be x and y

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

\sf{ x + y = 11......eq(1) }

\sf{ x - y = 19......eq(2) }

<u>Solving </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>get </u><u>:</u><u>-</u><u> </u>

\sf{ x + y = 11  }

\sf{ x = 11 - y ......eq(3 ) }

<u>Subsituting </u><u>eq(</u><u>3</u><u> </u><u>)</u><u> </u><u>in </u><u>eq</u><u>(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{ x - y = 19 }

\sf{ ( 11 - y) - y = 19 }

\sf{ 11 - y - y = 19 }

\sf{  11 - 2y = 19 }

\sf{ - 2y = 19 - 11  }

\sf{ - 2y = 8}

\sf{ y = 8/(-2) }

\sf{ y = - 4 }

\sf{\red{Thus ,\: the\: value\: of \: y = -4}}

<u>Now</u><u>, </u><u> </u><u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>3</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{ x = 11 - y  }

\sf{ x = 11 - (-4 )  }

\sf{ x = 11 + 4  }

\sf{ x = 15 }

Hence, The value of x and y are 15 and (-4) .

5 0
2 years ago
How big is the sun and how far is from the earth
ruslelena [56]
The sun is 432,690 mi big and the distance from the sun to earth is 92.75 million mi
4 0
3 years ago
I know the selected answer is correct but I'm not too sure how to get that answer.
Kryger [21]

\tt{ Hey \: there , \: Mr.Panda \: ! } ;)

♨\large{ \tt{ E \: X \: P \: L \: A \: N \: A \: T \: I\: O \: N}}:

⤻ Before solving the given question , you should know the answer of these questions :

✺How do you find the hypotenuse , perpendicular and base when the angle ( \theta \: , \alpha  \:  ,\beta ) is given ?

⇾ The longest side , which is the opposite side of right angle is the hypotenuse ( h ). There are two other sides , the opposite and the adjacent. The naming of these sides depends upon which angle is involved. The opposite is the side opposite the angle involved and it is called the perpendicular ( p ) . The adjacent us the side next to the angle involved ( buy not the hypotenuse ) and it is called the base ( b ).

☄ \large{ \tt{REMEMBER}} :

  • \bf{ \sin \theta =  \frac{opposite}{hypotenuse}  =  \frac{perpendicular}{hypotenuse}  }

  • \bf{ \cos\theta =  \frac{adjacent}{hypotenuse}  =  \frac{base}{hypotenuse}  }

  • \bf{ \tan \theta =  \frac{opposite}{adjacent}  =  \frac{perpendicular}{base}  }

In the above cases , \theta is taken as the angle of reference.

♪ Our Q/A part ends up here! Let's start solving the question :

❈ \large{ \tt{GIVEN}} :

  • Perpendicular ( p ) = ? , Hypotenuse ( h ) = 18 & base ( b ) = 16

✧ \large{ \tt{TO \: FIND} : }

  • Value of tan \theta

✎ \large{ \tt{SOLUTION}} :

Firstly , Finding the value of perpendicular ( p ) using Pythagoras theorem :

❃ \boxed{ \sf{ {h}^{2}  =  {p}^{2}  +  {b}^{2} }} [ Pythagoras theorem ]

\large{ ⇢ \sf{p}^{2}  +  {b}^{2}  =  {h}^{2} }

\large{⇢ \sf{ {p}^{2}  =  {h}^{2}  -  {b}^{2} }}

\large{ ⇢\sf{ {p}^{2}  =  {18}^{2}  -  {16}^{2} }}

\large{⇢ \sf{ {p}^{2}  = 324  - 256}}

\large{⇢ \sf{ {p}^{2}  = 68}}

\large{⇢ \sf{p =  \sqrt{68}}}

\large{ ⇢\sf{p =  \boxed{ \tt{2 \sqrt{17}}} }}

Okey, We found out the perpendicular i.e \tt{2 \sqrt{17}} . Now , We know :

❊ \large{ \sf{ \tan \theta} =  \frac{perpendicular}{base} }

\large {\tt{↬ \: tan \theta =  \frac{2 \sqrt{17} }{16}}}

\large{ \tt{ ↬ tan  \theta =  \frac{ \cancel{2} \:  \sqrt{17} }{ \cancel{16} \:  \: 8} }}

\large{ \tt{ ↬ \boxed{ \tt{tan \theta =  \frac{ \sqrt{17} }{8}}}}}

⟿ \boxed{ \boxed{ \tt{OUR\: FINAL \: ANSWER : \boxed{ \underline{ \bf{ \frac{ \sqrt{17} }{8}}}}}}}

۵ Yay! We're done!

♕ \large\tt{RULE \: OF \:SUCCESS }:

  • Never lose hope & keep on working ! ✔

ツ Hope I helped!

☃ Have a wonderful day / evening! ☼

# StayInAndExplore ☂

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3 0
3 years ago
Read 2 more answers
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