Answer:
Step-by-step explanation:
You could talk about how how or low it goes.
For instance: if the graph was going very high then suddenly low you could say that it was going very high until it suddenly dropped.
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Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Because she randomly added a 72 before the 78
<u>Answer:</u>
-10,1,19
<u>Step-by-step explanation:</u>
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x+y+z = 10 (Equation 1)
2y-x= 12 (Equation 2)
x-y+2z = 7 (Equation 3)
(Equation 2): -x = -2y+12
x = 2y-12 (Equation 4)
(Equation 1) - (Equation 3): 2y-z = 3
-z = -2y+3
z = 2y-3 (Equation 5)
Substitute (4) and (5) into (1)
x+y+z = 10
(2y-12)+y+(2y-3) = 10
5y-15 = 10
5y = 5
y=1
Substitute y=1 into (2)
2y-x= 12
2(1)-x= 12
2-x= 12
-x= 12-2
-x= 10
x= -10
Substitute y=1 and x=-10 into (1)
x+y+z = 10
-10+1+z = 10
z-9 = 10
z = 10+9
z = 19
Order: x = -10, y = 1, and z = 19
