Well, if the product has two decimal places and it is known that one factor has none, then the other factor must have two decimal places. Where would those two decimal places in the product come from otherwise?
Some examples:
1 * 34.44 = 34.44
2 * 17.22 = 34.44
In both cases, the non-integer factor had two decimal places.
Hope I helped, and let me know if you have any questions! :D
The answer would be 3x^2 I think.
Answer:
vertex at (1, -3)
Step-by-step explanation:
When x = 0
y² + 6y + 1 = 0
y² + 6y + = -1
y² + 6y + 9 = -1 + 9
(y + 3)² = 8 or (-y - 3)² = 8
y + 3 = √8 or -y - 3 = √8
y = - 3 +√8 or y = -3 - √8
(0, - 3 +√8) and (0, -3 - √8)
The mid point between these two is the average
y = ( - 3 + √8 + -3 - √8) / 2 = - 3
y² + 6y + 8x + 1 = 0
(-3)² + 6(-3) + 8x + 1 = 0
9 - 18 + 1 = -8x
- 8 = -8x
x = 1
tan 52 = 180/a; a = 180/tan 52
tan 43 = 180/b; b = 180/tan 43
you need to calculate, a-b = 180/tan 52 - 180/tan 43
substitute the value of tan thetas, and calculate it simply...
Answer:
1. Perpendicular
2. Isosceles
3. Never
Step-by-step explanation:
1. AC ⊥ BD because diameter of a square are perpendicular bisector of each other.
2. In Δ AOB , By using pythagoras : AB² = OA² + OB² .......( 1 )
In Δ COB , By using pythagoras : BC² = OC² + OB² ..........( 2 )
But, OA = OC because both are radius of same circle
So, by using equations ( 1 ) and ( 2 ), We get AB = BC ≠ AC
⇒ ABC is a triangle having two equal sides so ABC is an isosceles triangle.
3. The side can never be equal to radius of circle because the side of the square will be chord for the circle and in a circle chord can never be equal to its radius