Answer:
∠1 ≅ ∠2 ⇒ proved down
Step-by-step explanation:
#12
In the given figure
∵ LJ // WK
∵ LP is a transversal
∵ ∠1 and ∠KWP are corresponding angles
∵ The corresponding angles are equal in measures
∴ m∠1 = m∠KWP
∴ ∠1 ≅ ∠KWP ⇒ (1)
∵ WK // AP
∵ WP is a transversal
∵ ∠KWP and ∠WPA are interior alternate angles
∵ The interior alternate angles are equal in measures
∴ m∠KWP = m∠WPA
∴ ∠KWP ≅ ∠WPA ⇒ (2)
→ From (1) and (2)
∵ ∠1 and ∠WPA are congruent to ∠KWP
∴ ∠1 and ∠WPA are congruent
∴ ∠1 ≅ ∠WPA ⇒ (3)
∵ WP // AG
∵ AP is a transversal
∵ ∠WPA and ∠2 are interior alternate angles
∵ The interior alternate angles are equal in measures
∴ m∠WPA = m∠2
∴ ∠WPA ≅ ∠2 ⇒ (4)
→ From (3) and (4)
∵ ∠1 and ∠2 are congruent to ∠WPA
∴ ∠1 and ∠2 are congruent
∴ ∠1 ≅ ∠2 ⇒ proved
The complete question is as follows.
The equation a = 
can be used to determine the area , <em>a</em>, of a trapezoid with height , h, and base lengths, 
and 
. Which are equivalent equations?
(a) 
(b) 
(c) 
=
(d) 
(e) 
= h
Answer: (a) ; (d) ;
Step-by-step explanation: To determine :
a =
2a = (
)h
To determine h:
a =
2a = 
= h
To determine
a =
2a =
Checking the alternatives, you have that and = h, so alternatives <u>A</u> and <u>D</u> are correct.
The solution for this problem is:
The population is 500 times bigger since 8000/24 = 500. The population after t days is computed by:P(t) = P₀·4^(t/49)
Solve for t: 8000 = 8·4^(t/49) 1000 = 4^(t/49) log₄(1000) = t/49t = 49log₄(1000) ≅ 244 days
Answer:
D
Step-by-step explanation:
Differentiate the function and set it to zero:
f'(x) = -10x + 13
0 = -10x + 13
x = 1.3 seconds
Plug the x value into f(x):
f(1.3) = -5(1.3)^2 + 13(1.3) + 6 = 14.45 meters
Answer:1.75t + 3.50 ≤ 20
Step-by-step explanation:hope this helps :)
