Answer:
B
Step-by-step explanation:
hope this helps with the problems
Answer:
x=3
Step-by-step explanation:
Because the shape is a rhombus that means that WZ and XY are parallel
This means we can use the co-interior angles rule to conclude that ∠WZY and ∠XYZ add up to 180°
This means that ∠WZY is 44°
Because the shape in a rhombus ZX and WY are perpendicular
This means that ∠ZRY is a right angle (90°)
Because ZX and WY are perpendicular that means that ∠ZYX is split in half which makes ∠ZYW 68°
Because the angles in a triangle add up to 180°
(10x-8)+68+90=180
(10x-8)+158-158=180-158
10x-8+=22+8
10x/10=30/10
x=3
Answer:
y>2
y<x
Choice D
Step-by-step explanation:
y> 2 is shaded above the line
y< x is shaded below the line
Answer: The mat is 4.33 ft high off the ground.
Step-by-step explanation:
Since we have given that
Angle of elevation with the first triangle = 30°
Angle of elevation with the second triangle = 60°
Length at which gymnastics mat extends across the floor = 5 feet
so, As shown in the figure:
We need to find the height of the mat off the ground.
If CD = 5 ft,
Let, AB = y, DC = x.
In Δ ABC,

Similarly, in Δ ACD,

Hence, the mat is 4.33 ft high off the ground.