Answer:
(x-2), (x+2), (3x-5)
Step-by-step explanation:
Factors of 3: ±1, ±3
Factors of 20: ±1, ±2, ±4, ±5, ±10, ±20
Possible factors of the polynomial: ±1, ±2, ±3, ±4, ±5, ±10, ±20, .... (there's a lot more but you probably do not need to list them all)
Pick a number to divide the polynomial by (I picked 2)
(3x³-5x²-12x+20)÷(x-2) = 3x²+x-10
So (x-2) is a factor of f(x) = 3x³-5x²-12x+20
Factor 3x²+x-10 = (3x-5)(x-2) these are the other factors of f(x) = 3x³-5x²-12x+20
ANSWER
and
EXPLANATION
Line s has equation:
y=x-5
And line t has equation:
We equate the two equations to get:
Multiply through by 4
7x=28
Answer:
ΔPTS≅ΔRTA by AAS axiom of congruency
Step-by-step explanation:
Consider ΔPQA and ΔRQS
∠PQA=∠RQS (Vertically Opposite Angles)
∠QAP=∠QSR (Complementary of two equal angles, ∠RAT and∠PST)
Due to angle sum property of a triangle, we come to the conclusion that
∠APQ=∠SRQ
Consider ΔPTS and ΔRTA
TA=TS (Given)
∠RAT=∠PST(Given)
∠APQ=∠SRQ (Proved above)
Therefore, ΔPTS≅ΔRTA by AAS axiom of congruency.
Answer:
We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.
We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :
1 - We first group terms with the same variable :
(x²+8x) + (y²+22y) + 37 = 0
2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)
(x²+8x) + (y²+22y) = - 37
3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.
x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²
y²+22y = (y²+22y+11²)-11² = (y+11)²-11²
4 - We plug the new values inside our equation :
(x+4)² - 4² + (y+22)² - 11² = -37
(x+4)² + (y+22)² = -37+4²+11²
(x+4)²+(y+22)² = 100
5 - We re-write in standard form :
(x-(-4)²)² + (y - (-22))² = 10²
And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)
