If you multiply by 24 you should get $864
second-period class average:
55+70+450+480+170+270+95 = 1590
1590/20 = 79.5
sixth-period class average:
65+225+480+595+270+190 = 1825
1825/20 = 91.25
on average, students in the sixth period class scored higher
Answer:
x = 41°, y = 139°
Step-by-step explanation:
The given parameters are;
Line <em>m</em> is parallel to line <em>n</em> and lines <em>m</em> and <em>n</em> have a common transversal
The corresponding angles formed by the common transversal to the two parallel lines are 41° on line <em>m</em> and <em>x°</em> on line <em>n</em>
Therefore, x° = 41° by corresponding angles formed between on two parallel lines by a common transversal are equal
x° and y° are linear pair angles and they are, supplementary
∴ x° + y° = 180°
∴ x° + y° = 41° + y° = 180°
y° = 180° - 41° = 139°
y° = 139°.
Hello!
∠GCE = 14° (corresponding angles)
∠GEC = 180° - 78° - 14° (sum of angles in a triangle)
∠GEC = 88°
4x + 88° = 180°
4x = 180° - 88°
4x = 92°
x = 23°
We know that the points at which the parabola intersects the x axis are
(-5,0) and (1,0)
So the extent between these two points would be the base of the triangle
lets find the length of the base using the distance formula
![\sqrt{[(-5-1)^{2}+(0-0)^{2} ]}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%5B%28-5-1%29%5E%7B2%7D%2B%280-0%29%5E%7B2%7D%20%5D%7D%20%20)
the base b=6
We will get the height of the triangle when we put x=0 in the equation
y=a(0+5)(0-1)
y=-5a
so height = -5a (we take +5a since it is the height)
We know that the area of the triangle =
× 6 × (5a) = 12
15a=12
a= 
