Answer:
∠ U = 22.62°
Step-by-step explanation:
We can use the trigonometric expression to find ∠U by using the ratio:
sin(θ) = opposite/ hypotenuse
=> sin ∠U = 7.5/ 19.5
=> ∠ U = 
( 7.5/ 19.5)
∴ ∠ U = 22.619° ≈ 22.62°
Angles formed by the segment 
in the triangles ΔWXZ, and ΔXYZ, are equal and the given corresponding sides are proportional.
- The option that best completes the proof showing that ΔWXZ ~ ΔXYZ is; <u>16 over 12 equals 12 over 9</u>
Reasons:
The proof showing that ΔWXZ ~ ΔXYZ is presented as follows;
Segment is perpendicular to segment 
∠WZX and ∠XZY are right angles by definition of perpendicular to
∠WZX in ΔWXZ = ∠XZY in ΔXYZ = 90° (definition)
Therefore;
, which gives, 
Given that two sides of ΔWXZ are proportional to two sides of ΔXYZ, and
that the included angles between the two sides, ∠WZX and ∠XZY are
congruent, the two triangles, ΔWXZ and ΔXYZ are similar by Side-Angle-
Side, SAS, similarity postulate.
The option that best completes the proof is therefore;
- which is; <u>16 over 12 equals 12 over 9</u>
Learn more about the SAS similarity postulate here:
brainly.com/question/11923416
The area of a trapezoid can be computed as:
A=(b1+b2)h/2, where b1 and b2 ar the bases, h is the height.
With the provided numerical values:
50=(3+7)h/2
50=10h/2=5h
Answer: h=10inches
Step-by-step explanation:
problem 1.
1. 6y = 30
2. y = 5
3. 3x + 2(5) = 16
4. 3x+10=16
5. 3x=6
6. x=2
solution: (2,5)
problem 2.
1. x=7
2. 4(2)-2y=18
3. 28-2y=18
4. -2y= -10
5. y=5
solution (7,5)
problem 3
10x=10
x=1
7(1) +y =-2
y=-9
(1,9)
problem 4
0= -6
no solution
problem 5
0=0
Infinite many
Answer:
v_top = 2400 mi/hr
v_w = 400 mi/h
Step-by-step explanation:
Given:
- Total distance D = 4800 mi
- Headwind journey time taken t_up= 3 hr
- Tailwind journey time taken t_down = 2 hr
Find:
Find the top speed of Luke's snow speeder and the speed of the wind.
Solution:
- The speed of Luke v_l is in stationary frame is given by:
v_l = v_w + v_l/w
Where,
v_w: Wind speed
v_l/w: Luke speed relative to wind.
- The top speed is attained on his returned journey with tail wind. We will use distance time relationship to calculate as follows:
v_top = D / t_down
v_top = 4800 / 2
v_top = v_down = 2400 mi/hr
- Similarly his speed on his journey up with head wind was v_up:
v_up = D / t_up
v_up = 4800 / 3
v_up = 1600 mi/hr
- Now use the frame relations to find the wind speed v_w:
v_down = v_w + v_l/w
v_up = -v_w + v_l/w
- Solve equations simultaneously:
2400 = v_w + v_l/w
1600 = -v_w + v_l/w
4000 = 2*v_l/w
v_l/w = 2000 mi/h
v_w = 400 mi/h
