<u>Finding x:</u>
We know that the diagonals of a rhombus bisect its angles
So, since US is a diagonal of the given rhombus:
∠RUS = ∠TUS
10x - 23 = 3x + 19 [replacing the given values of the angles]
7x - 23 = 19 [subtracting 3x from both sides]
7x = 42 [adding 23 on both sides]
x = 6 [dividing both sides by 7]
<u>Finding ∠RUT:</u>
We can see that:
∠RUT = ∠RUS + ∠TUS
<em>Since we are given the values of ∠RUS and ∠TUS:</em>
∠RUT = (10x - 23) + (3x + 19)
∠RUT = 13x - 4
<em>We know that x = 6:</em>
∠RUT = 13(6)- 4
∠RUT = 74°
Answer:
5,600 m^2
Step-by-step explanation:
a 1cm= 0.01m
560000(0.01)=5600
Answer:
The distance between the base of the tree and the flower is 9m
Step-by-step explanation:
Here, we have to paint a picture.
The flower is on the ground, the height of the tree is 12m.
The distance from the nest to the flower on the floor is 15m
Indisputably, what we have is a right angled triangle, with the height being 12m, the length of the hypotenuse being 15 and we are asked to calculate the adjacent which represents the distance from the base of the tree to the flower
To get this distance, we simply apply the Pythagoras’ theorem which states that the square of the hypotenuse(longest side of the triangle) is equal to the sum of the squares of the other two sides.
Thus mathematically, we know that our hypotenuse is 15m and the height is 12m
The length we are to calculate is the adjacent and it is equal to;
15^2 - 12^2
= 225 - 144
= 81
The length is thus
√(81) = 9m
Please check attachment for a diagrammatic picture of the triangle
Not on your list, but an easy way is
.. a) swap coefficients of x and y, negating one. (Now you have 2x -3y.)
.. b) set any constant term to zero (now you have 2x -3y = 0)
.. c) translate the line to the point (5, 2) by substituting x ⇒ x-5, y⇒ y-2
2(x -5) -3(y -2) = 0
The way you've been taught, selection C is the proper choice.
I believe the correct answer from the choices listed above is the last option. The last option clearly describes the illustration of the construction of a perpendicular to a line from a point on the line where you start on a point in the line. Using an arbitrary radius, draw arcs intersecting the line <span> at two points. </span>
Hope this answers the question. Have a nice day.