CENTROID is the term used to describe the point where the three medians of a triangle intersect.
Centroid is the center of gravity of the triangle. It is one of the triangle's point of concurrency. Aside from "center of gravity", it is also known as "center of mass" and "barycenter".
Centroids are always inside the triangle. Since, the centroid is where three medians intersect, each median divides the triangle into two smaller triangles of equal area. The centroid is exactly two-thirds the way along each median.
Other types of triangle centers aside from Centroids are:
1) incenter
2) circumcenter
3) orthocenter
The numbers are: "9" and "12" .
___________________________________
Explanation:
___________________________________
Let: "x" be the "first number" ; AND:
Let: "y" be the "second number" .
___________________________________
From the question/problem, we are given:
___________________________________
2x + 5y = 78 ; → "the first equation" ; AND:
5x − y = 33 ; → "the second equation" .
____________________________________
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 " .
_____________________________________________
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".
____________________________________________________
Now, let us take "the first equation" ; which is:
" 2x + 5y = 78 " ;
_______________________________________
We have our obtained value; " y = 5x − 33 " .
_______________________________________
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" ;
________________________________________________
Take the "first equation":
________________________________________________
→ " 2x + 5y = 78 " ; and write as:
________________________________________________
→ " 2x + 5(5x − 33) = 78 " ;
________________________________________________
Note the "distributive property of multiplication" :
________________________________________________
a(b + c) = ab + ac ; AND:
a(b − c) = ab − ac .
________________________________________________
So; using the "distributive property of multiplication:
→ +5(5x − 33) = (5*5x) − (5*33) = +25x − 165 .
___________________________________________________
So we can rewrite our equation:
→ " 2x + 5(5x − 33) = 78 " ;
by substituting the: "+ 5(5x − 33) " ; with: "+25x − 165" ; as follows:
_____________________________________________________
→ " 2x + 25x − 165 = 78 " ;
_____________________________________________________
→ 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.
_____________________________________________________
→ Rewrite the equation as:
_____________________________________________________
→ " 27x − 165 = 78 " ;
Now, add "165" to EACH SIDE of the equation; as follows:
→ 27x − 165 + 165 = 78 + 165 ;
→ to get: 27x = 243 ;
_____________________________________________________
Now, divide EACH SIDE of the equation by "27" ; to isolate "x" on one side of the equation ; and to solve for "x" ;
_____________________________________________________
27x / 27 = 243 / 27 ;
→ to get: x = 9 ; which is "the first number" .
_____________________________________________________
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);
______________________________
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" .
____________________________________________
To check our work:
_______________________
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!
________________________
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!
_____________________________________
So, these answers do make sense!
______________________________________
Answer:
?= 2
X=1
Step-by-step explanation:
3(1+1)<9
3×2<9
6<9
Pls mark brainliest.
Answer:
B
Step-by-step explanation:
Given
6p³ - 12p² + 9p ← factor out 3p from each term
= 3p(2p² - 4p + 3) → B
Answer:
C, -2
Step-by-step explanation:
-4x + 7 = 2y - 3
-4x + 10 = 2y
-2y = 4x - 10
y = -2x + 5