Answer:
1. UW // TX
2. VX // UY
3. UW ≅ TY ≅ YX
4. YW = 
TV
5. TX = 2 UW
6. ∠TXV ≅∠WUY
Step-by-step explanation:
The line segment joining the midpoint of two sides of a triangle is parallel to the third side and equal to half its length
In Δ XVT
∵ U is the midpoint of VT
∵ W is the midpoint of VX
∵ XT is the 3rd side of the triangle
→ By using the rule above
∴ UW // TX ⇒ (1)
∴ UW = TX
→ Multiply both sides by 2
∴ 2 UW = TX
∴ TX = 2 UW ⇒ (5)
∵ Y is the midpoint of TX
∴ TY = YX = TX
∵ UW = TX
∴ UW ≅ TY ≅ YX ⇒ (3)
∵ U is the midpoint of VT
∵ Y is the midpoint of XT
∵ VX is the 3rd side of the triangle
→ By using the rule above
∴ UY // VX
∴ VX // UY ⇒ (2)
∴ UY = VX
∵ W is the midpoint of VX
∵ Y is the midpoint of XT
∵ TV is the 3rd side of the triangle
→ By using the rule above
∴ YW // TV
∴ YW = TV ⇒ (4)
∵ 2 Δs UYW and XVT
∵ UY = XV
∵ YW = VT
∵ WU = TX
∴ 
= 
= 
=
→ By using the SSS postulate of similarity
∴ ∠TXV ≅∠WUY ⇒ (6)
Answer:
D
Step-by-step explanation:
12-(-3)=15
Answer:
1/6
Step-by-step explanation:
Given:
- Length of the trough: 9 ft
=> The volume of the trough: V =
* (b * h) (1)
- An isosceles right triangle with hypotenuse 2 feet
=> the other two sides of the triangle is:
= tan(45 degrees) = h/(b/2)
<=> b = 2h substitute in (1), we have:
V = *(2h *h) = 9
Take derivative of volume with respect to time to find equation for rate of filling the trough
dV/dt = 2 * 9 *h dh/dt = 18h dh/dt
<=> dh/dt = dV/dt /(18h)
As we know that, dV/dt = 2
So, dh/dt = 2 / 18h = 1/9h
<=> V = t * rate = 2 * 2 = 4
But V = 9
<=> 9 = 4
<=> h = 2/3
The rate is the height h feet of the water in the trough changing 2 minutes after the water begins to flow:
dh/dt = 1/(9h) = 1/(9 * 2/3) = 1/6
Answer:
5/4
Step-by-step explanation:
y=mx+b
m = slope
5/4= slope
Answer:
A. 240/7
Step-by-step explanation:
10^4x/24 * 10^3x/24 = 10^10
10^7x/24 = 10^10
7x/24 = 10
x = 240/7
