How to find rapidly the coordinates of Q:
since Q is the center of gravity of the triangle ABC, so we have the following vector relationship
vecQA +vecQB +vecQC =<span>vec0
</span><span>vecQA=(x-3, y+2)
</span><span>vecQB=(x-1, y+5)
</span><span>vecQC=(x-7, y+5)
</span><span>vec0=(0, 0)
</span>
so, vecQA +vecQB +vecQC =<span>vec0 is equivalent to
</span>x-3 +x-1+x-7 =0, and y+2+y+5+<span>y+5=0 so 3x-11=0 implies x=11/3
</span><span>and 3y+12=0 implies y=-12/3
finally the </span><span>the coordinates of point Q is (11/3, -4)</span><span>
</span>
From the diagram, all three angles are 90° (Diagonals of a Rhombus)
Angle 1 = 9x = 90
⇒ 9x = 90
x = 90/9
x = 10
Angle 2 = y + x = 90
⇒ y + x = 90
since x = 10,
y + 10 = 90
y = 90 - 10
y = 80
Angle 3 = 15 z = 90
⇒ 15z = 90
z = 90/15
z = 6
Answer is Option C
<span>C. x = 10, y = 80, z = 6</span>
Answer:
-27
Step-by-step explanation:
ok so we have (-9) in parentheses right? And positive 3 outside of parentheses so we know 9 x 3 = 27 BUT we have a negative in front of 9 so we just put -27.
Hope this helps :D ~P.s- pls could u give brainliest to meh TYSM!
Solving equations simple
4n + 3 =11
Y = 4
y intercept is 4
The line would be horizontal and it would intersect the y axis at the point (0,4)
