If the side of square P is x and the side of square Q is y, we have x+y=26. In addition, 3*the perimeter of P+8*4y=3*4x+32y=12x+32y=492.
We have
x+y=26
12x+32y=492
Multiplying the first equation by -12 and adding it to the second, we get
20y=180 and dividing by 20 we get y=9. Plugging it into the first equation, we get 26-9=17=x. Since the area of square P is x^2 and Q is y^2,
x^2+y^2=19^2+7^2=410=the sum of the squares
Answer:
The second option, 3x^2(x+3)(4x+1)
Step-by-step explanation:
Factor out the common term, which is 3x^2
This then becomes 3x^2(4x^2+13x+3)
Factor the inside term and it becomes
(3x^2)(x+3)(4x+1)
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
Answer:
The clock hand made a 90 degree turn. It is not a 3/4 fraction.
Answer:
2.3% as a decimal is 0.023