The line contains point (2, 0) and (-5, 4).
y - 0 = (4 - 0)/(-5 - 2) (x - 2)
y = -4/7 (x - 2)
7y = -4(x - 2)
7y = -4x + 8
4x + 7y = 8
10 because 45x3=135 205-125=70 70 divided by 7 equals 10
im sorry but i cant seem to open your image?
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
In order for the ball to be used in the game, it must be able to meet the minimum and maximum weight requirements. These are the limits of the weight of the ball. If it exceeds the maximum limit of 16 ounces or below the minimum limit of 14 ounces, the ball will not be approved.
So, by adding 1.5 ounces, that would mean that the initial weight of the ball did not reach the minimum limit. The initial weight of the ball, denoted as x, have two possible values. The first value is the initial weight plus added with 1.5 ounces would reach 14 ounces.
x + 1.5 = 14
x = 12.5 ounces
The other scenario is when the initial weight is added to reach the maximum requirement of 16 ounces
x + 1.5 = 16
x = 14.5 ounces
From both answer, we could conclude that the initial weight has to be 12.5 ounces. If the initial weight were 14.5 ounces to begin with, there should be no need for air. It cans till be approved to be used.
