The perimeter of RSTU will be C.22m if you look at both of them u will see they are both the same and tell you how long the sides are,........hope i was right:))
Answer: (0,0),(0,4), and (5,0)
Step-by-step explanation:
It is easy to find lengths of horizontal and vertical segments and distances from (0,0) so always place one vertex at the origin and one or more sides on an axis.
This answer has an x intercept and y intercept.
Answer:
[Vertex form]
Step-by-step explanation:
Given function:
We need to find the vertex form which is.,
where 
represents the co-ordinates of vertex.
We apply completing square method to do so.
We have
First of all we make sure that the leading co-efficient is =1.
In order to make the leading co-efficient is =1, we multiply each term with -3.
Isolating 
and 
terms on one side.
Subtracting both sides by 15.
In order to make the right side a perfect square trinomial, we will take half of the co-efficient of term, square it and add it both sides side.
square of half of the co-efficient of term = 
Adding 36 to both sides.
Since 
is a perfect square of 
, so, we can write as:
Subtracting 21 to both sides:
Dividing both sides by -3.
[Vertex form]
The numbers are: "9" and "12" .
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Explanation:
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Let: "x" be the "first number" ; AND:
Let: "y" be the "second number" .
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From the question/problem, we are given:
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2x + 5y = 78 ; → "the first equation" ; AND:
5x − y = 33 ; → "the second equation" .
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From "the second equation" ; which is:
" 5x − y = 33" ;
→ Add "y" to EACH side of the equation;
5x − y + y = 33 + y ;
to get: 5x = 33 + y ;
Now, subtract: "33" from each side of the equation; to isolate "y" on one side of the equation ; and to solve for "y" (in term of "x");
5x − 33 = 33 + y − 33 ;
to get: " 5x − 33 = y " ; ↔ " y = 5x − 33 " .
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Note: We choose "the second equation"; because "the second equation"; that is; "5x − y = 33" ; already has a "y" value with no "coefficient" ; & it is easier to solve for one of our numbers (variables); that is, "x" or "y"; in terms of the other one; & then substitute that value into "the first equation".
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Now, let us take "the first equation" ; which is:
" 2x + 5y = 78 " ;
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We have our obtained value; " y = 5x − 33 " .
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We shall take our obtained value for "y" ; which is: "(5x− 33") ; and plug this value into the "y" value in the "first equation"; and solve for "x" ;
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Take the "first equation":
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→ " 2x + 5y = 78 " ; and write as:
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→ " 2x + 5(5x − 33) = 78 " ;
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Note the "distributive property of multiplication" :
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a(b + c) = ab + ac ; AND:
a(b − c) = ab − ac .
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So; using the "distributive property of multiplication:
→ +5(5x − 33) = (5*5x) − (5*33) = +25x − 165 .
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So we can rewrite our equation:
→ " 2x + 5(5x − 33) = 78 " ;
by substituting the: "+ 5(5x − 33) " ; with: "+25x − 165" ; as follows:
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→ " 2x + 25x − 165 = 78 " ;
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→ Now, combine the "like terms" on the "left-hand side" of the equation:
+2x + 25x = +27x ;
Note: There are no "like terms" on the "right-hand side" of the equation.
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→ Rewrite the equation as:
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→ " 27x − 165 = 78 " ;
Now, add "165" to EACH SIDE of the equation; as follows:
→ 27x − 165 + 165 = 78 + 165 ;
→ to get: 27x = 243 ;
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Now, divide EACH SIDE of the equation by "27" ; to isolate "x" on one side of the equation ; and to solve for "x" ;
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27x / 27 = 243 / 27 ;
→ to get: x = 9 ; which is "the first number" .
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Now; Let's go back to our "first equation" and "second equation" to solve for "y" (our "second number"):
2x + 5y = 78 ; (first equation);
5x − y = 33 ; (second equation);
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Start with our "second equation"; to solve for "y"; plug in "9" for "x" ;
→ 5(9) − y = 33 ;
45 − y = 33;
Add "y" to each side of the equation:
45 − y + y = 33 + y ; to get:
45 = 33 + y ;
↔ y + 33 = 45 ; Subtract "33" from each side of the equation; to isolate "y" on one side of the equation ; & to solve for "y" ;
→ y + 33 − 33 = 45 − 33 ;
to get: y = 12 ;
So; x = 9 ; and y = 12 . The numbers are: "9" and "12" .
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To check our work:
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1) Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;
→ 5x − y = 33 ; → 5(9) − 12 =? 33 ?? ; → 45 − 12 =? 33 ?? ; Yes!
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2) Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;
→ 2x + 5y = 78 ; → 2(9) + 5(12) =? 78?? ; → 18 + 60 =? 78?? ; Yes!
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So, these answers do make sense!
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The common denominator would be 18. So, 5/6 becomes 15/18 and 7/9 becomes 14/18
