Answer:you can look that up and it will give you the answer on google or photo math
Step-by-step explanation:
Answer:
Here is the full proof:
AC bisects ∠BCD Given
∠CAB ≅ ∠CAD Definition of angle bisector
DC ⊥ AD Given
∠ADC = 90° Definition of perpendicular lines
BC ⊥ AB Given
∠ABC = 90° Definition of perpendicular lines
∠ADC ≅ ∠ABC Right angles are congruent
AC = AC Reflexive property
ΔCAB ≅ ΔCAD SAA
BC = DC CPCTC
<span>y=1/3 x - 2
3y = x - 6
x - 3y = 6 ...this is standard form</span>
Answer:
a. attached graph; zero real: 2
b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
Step-by-step explanation:
p(x) = x³ + 4x² + 6x - 36
a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.
p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0
b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.
2 | 1 4 6 -36
1 6 18 0
x² + 6x + 18 = 0
Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²
√Δ = 6i
x = -6±6i/2 = 2(-3±3i)/2
x' = -3-3i
x" = -3+3i
p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)
c. the solutions are 2, -3-3i and -3+3i
Answer:
a2 – b2 = (a – b)(a + b)
(a + b)2 = a2 + 2ab + b2
a2 + b2 = (a + b)2 – 2ab
(a – b)2 = a2 – 2ab + b2
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
(a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
(a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
a4 – b4 = (a – b)(a + b)(a2 + b2)
a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4
