Answer:
H
Step-by-step explanation:
3:2 IS EQUAL TO 3/2
Let x represent the smaller. Then x+1 is the greater of the two.
... x+1 = 2x +20
... 0 = x + 19 . . . . . subtract x+1
... x = -19
Your two integers are -19 and -18.
Answer:
1250 m²
Step-by-step explanation:
Let x and y denote the sides of the rectangular research plot.
Thus, area is;
A = xy
Now, we are told that end of the plot already has an erected wall. This means we are left with 3 sides to work with.
Thus, if y is the erected wall, and we are using 100m wire for the remaining sides, it means;
2x + y = 100
Thus, y = 100 - 2x
Since A = xy
We have; A = x(100 - 2x)
A = 100x - 2x²
At maximum area, dA/dx = 0.thus;
dA/dx = 100 - 4x
-4x + 100 = 0
4x = 100
x = 100/4
x = 25
Let's confirm if it is maximum from d²A/dx²
d²A/dx² = -4. This is less than 0 and thus it's maximum.
Let's plug in 25 for x in the area equation;
A_max = 25(100 - 2(25))
A_max = 1250 m²
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
Answer:
It could either be x * 3.25 or x + 18
Step-by-step explanation:
8 * 3.25 = 26
8 + 18 = 26
