Answer:
The correct answer is diagonals are equal and bisect each other.
Step-by-step explanation:
A rectangle is a quadrilateral with opposites sides equal in length and all interior angles equal to 90°. Opposite sides are parallel and adjacent sides are perpendicular. Diagonals in a rectangle are equal in length and they bisect each other. If we want to draw a circumcircle, then the point of intersection of these diagonals give the circumcenter.
Answer:
x^2 -12x+20
Step-by-step explanation:
If the zeros of the functions are at 10 and 2
(x-10) (x-2)
FOIL
x^2 -2x-10x+20
Combine like terms
x^2 -12x+20
Answer:
x = (-29)/5
Step-by-step explanation:
Solve for x:
(2 (x - 5))/3 = (3 (x + 1))/2
Multiply both sides by 6:
(6×2 (x - 5))/3 = (6×3 (x + 1))/2
6/3 = (3×2)/3 = 2:
2×2 (x - 5) = (6×3 (x + 1))/2
6/2 = (2×3)/2 = 3:
2×2 (x - 5) = 3×3 (x + 1)
2×2 = 4:
4 (x - 5) = 3×3 (x + 1)
3×3 = 9:
4 (x - 5) = 9 (x + 1)
Expand out terms of the left hand side:
4 x - 20 = 9 (x + 1)
Expand out terms of the right hand side:
4 x - 20 = 9 x + 9
Subtract 9 x from both sides:
(4 x - 9 x) - 20 = (9 x - 9 x) + 9
4 x - 9 x = -5 x:
-5 x - 20 = (9 x - 9 x) + 9
9 x - 9 x = 0:
-5 x - 20 = 9
Add 20 to both sides:
(20 - 20) - 5 x = 20 + 9
20 - 20 = 0:
-5 x = 9 + 20
9 + 20 = 29:
-5 x = 29
Divide both sides of -5 x = 29 by -5:
(-5 x)/(-5) = 29/(-5)
(-5)/(-5) = 1:
x = 29/(-5)
Multiply numerator and denominator of 29/(-5) by -1:
Answer: x = (-29)/5
Here we are given with a triangle with smaller triangles formed due to the altitude on AC. Given:
- AB = 6
- BC = 8
- <ABC = 90°
- BD ⊥ AC
- <ABD =
We have to find the value for sin
So, Let's start solving....
In ∆ADB and ∆ABC,
- <A = <A (common)
- <ABC = <ADB (90°)
So, ∆ADB ~ ∆ABC (By AA similarity)
The corresponding sides will be:

We know the value of AB and to find AC, we can use Pythagoras theoram that is:
AC = √6² + 8²
AC = 10
Coming back to the relation,


In ∆ADB, we have to find sin
which is given by perpendicular/base:

Plugging the values of AD and AB,

Simplifying,

And this is our final answer.....
Carry On Learning !