Answer:
x = 10
Step-by-step explanation:
Solve for x:
-8 (8 - x) = (4 (x + 10))/5
Multiply both sides by 5:
-8×5 (8 - x) = (5×4 (x + 10))/5
(5×4 (x + 10))/5 = 5/5×4 (x + 10) = 4 (x + 10):
-8×5 (8 - x) = 4 (x + 10)
5 (-8) = -40:
-40 (8 - x) = 4 (x + 10)
Expand out terms of the left hand side:
40 x - 320 = 4 (x + 10)
Expand out terms of the right hand side:
40 x - 320 = 4 x + 40
Subtract 4 x from both sides:
(40 x - 4 x) - 320 = (4 x - 4 x) + 40
40 x - 4 x = 36 x:
36 x - 320 = (4 x - 4 x) + 40
4 x - 4 x = 0:
36 x - 320 = 40
Add 320 to both sides:
36 x + (320 - 320) = 320 + 40
320 - 320 = 0:
36 x = 40 + 320
40 + 320 = 360:
36 x = 360
Divide both sides of 36 x = 360 by 36:
(36 x)/36 = 360/36
36/36 = 1:
x = 360/36
The gcd of 360 and 36 is 36, so 360/36 = (36×10)/(36×1) = 36/36×10 = 10:
Answer: x = 10
An outside angle equals the sum of the two opposite inside angles.
So you have:
(5x-35) = 2x +58
Solve for X then you can calculate Angle B:
(5x-35) = 2x +58
Subtract 2x from both sides:
3x -35 = 58
Add 35 to both sides:
3x = 93
Divide both sides by 3:
X = 93 /3
x = 31
Now for Angle B, replace x with 31:
Angle B = 2x = 2(31) = 62 degrees.
7) answer-6t
t=toms cans collected
8) answer- 5.5
8.3-2.8=5.5
9) answer-14
56/4=14
Given
26% = several simulations of a football player playing two games
Find the percentage of the simulations in each of the two games
2x = .26
2x = .26
---- -----
2 2
x = .13 * 100
x= 13
The answer is 13% of simulations would the football player be most likely to get a touchdown in each of the two games.
Answer:
Hence proved triangle ADE ≅ triangle BCE by Side Angle Side congruent property.
Step-by-step explanation:
Given:
AD ⊥ AB 
CD
BC ⊥ AB CD
AD = BC
∴ ∠ A = ∠ B = ∠ C = ∠ D =90°
∠ EDC = ∠ ECD
Solution
∠ C = ∠ BCE + ∠ ECD⇒ equation 1
∠ D = ∠ ADE + ∠ EDC⇒ equation 2
∠ C = ∠ D (given)
Substituting equation 1 and 2 in above equation we get
∠ BCE + ∠ ECD = ∠ ADE + ∠ EDC
But ∠ EDC = ∠ ECD (given)
∴ ∠ ADE = ∠ BCE
ED = EC (∵ base angles are same triangle is isosceles triangle)
Now, In Δ ADE and Δ BCE
AD =BC
∠ ADE = ∠ BDE
ED = EC
∴ By Side Angle Side congruent property
Δ ADE ≅ Δ BCE
