True
I have notes on this so im pretty sure
-4.46cm
Step-by-step explanation:
The horizontal midpoint of the page is;
The horizontal midpoint of the image is given as;
For the image to be centered, these to midpoints must be on the same point in the axis.
If we take the leftmost edge of the paper to be point zero on the axis, then, the distance accommodated by the paper is 10.795cm. The distance that goes beyond this leftmost edge is computed as;
On the axis, this is on the negative side. Therefore, the horizontal position for the left edge of the image that Joelle should use to center the image is -4.46cm.
Answer:
1) x = 40° & y = 50°
2) x = 100° & y = 80°
Step-by-step explanation:
1)
ABCD is a cyclic quadrilateral (∵ all the vertices of the quadrilateral are touching on the circle). As it is a cyclic quadrilateral , sum of it's opposite angles will be 180°.
⇒ ∠ADC + ∠ABC = 180°
⇒ 130° + y = 180°
⇒ y = 180 - 130 = 50°
In ΔABC ,
∠ACB = 90° (∵ AB is the diameter of the circle and a diameter subtends an angle of 90° on any point on circle.)
Using angle sum property of triangle ,
∠ABC + ∠ACB + ∠CAB = 180°
⇒ y + 90° + x = 180°
⇒ x + 50° + 90° = 180°
⇒ x + 140° = 180°
⇒ x = 180 - 140 = 40°
2)
ΔABC is an isosceles triangle (∵AB = AC). As it is an isosceles triangle , it's base angles will be equal. So , ∠ABC = ∠ACB = 50°
Using angle sum property of triangle ,
∠ABC + ∠ACB + ∠BAC = 180°
⇒ 50° + 50° + y = 180°
⇒ y + 100° = 180°
⇒ y = 180 - 100 = 80°
ABEC is a cyclic quadrilateral (∵ all the vertices of the quadrilateral are on the circle.). As it is a cyclic quadrilateral , sum of it's opposite angles will be 180°.
⇒ ∠BAC + ∠BEC = 180°
⇒ y + x = 180°
⇒ x + 80° = 180°
⇒ x = 180 - 80 = 100°
The answer you want is 20!
Answer: 784 weeks
Step-by-step explanation:
