A number can be represented by n. So n/15≤450 then multiply both sides by 15 to get n by itself n≤6,750
Hope this helped!!! :)
Answer:
4x-13 = 2x+9 ( i think it was exterior angle property i dont remember but its true)
so, solving that eq,
4x-2x = 9+13 (grouping variables and constants)
2x=22
x=11
also, 2x+9 and 3y+2 are on a transversal, and are supplementary angles (add up to 180°)
so, solving for x in 2x+9,
2(11)+9 = 22+9 = 31°
now, 31 + 3y+2 = 180
3y = 180 - 33
y = 147/3 = 49
(hoping you understood the calculations :)
Answer:
-6, -2, 1, 8
Step-by-step explanation:
Ok, so user says that it should be solve for vertex not vertex form
(x,y)
to find the vertex of
y=ax^2+bx+c
the x value of the vertex is -b/2a
the y value is found by plugging in the x value for the vertex back into the original equation and evaluating
y=-2x^2-12x-28
a=-2
b=-12
xvalue of vertex is -(-12)/(2*-2)=12/-4=-3
x value of vertex is -3
plug backin for x
y=-2x^2-12x-28
y=-2(-3)^2-12(-3)-28
y=-2(9)+36-28
y=-18+8
y=-10
yvalue is -10
x value is -3
vertex is (-3,-10)
Answer:
See below.
Step-by-step explanation:
This is how you prove it.
<B and <F are given as congruent.
This is 1 pair of congruent angles for triangles ABC and GFE.
<DEC and <DCE are given as congruent.
Using vertical angles and substitution of transitivity of congruence of angles, show that angles ACB and GEF are congruent.
This is 1 pair of congruent angles for triangles ABC and GFE.
Now you need another side to do either AAS or ASA.
Look at triangle DCE. Using the fact that angles DEC and DCE are congruent, opposite sides are congruent, so segments DC and DE are congruent. You are told segments DF and BD are congruent. Using segment addition postulate and substitution, show that segments CB and EF are congruent.
Now you have 1 pair of included sides congruent ABC and GFE.
Now using ASA, you prove triangles ABC and GFE congruent.
