Answer:
y=3/2x+1
Step-by-step explanation:
y-4= -⅔(x-6)
y-4=-⅔x+4
y=-⅔x+4+4
(equation of line 1) y= -⅔x+8 gradient= -⅔
(line 2)gradient=3/2
note* the gradients of perpendicular lines multiplied result to -1
gradient=<u>y²-y²</u>
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>x²-x¹
<u>3</u><u> </u>=<u>y</u><u>+</u><u>2</u>
<u> </u><u> </u>2. x+2
multiply both sides by 2(x+2)to remove the denominators
3(x+2)=2(y+2)
3x+6=2y+4
3x+6-4=2y
3x+2=2y
divide all sides by 2
3/2x+1=y
y=3/2x+1
Answer:
Step-by-step
i) a² - 2ab + b² = (a + b)²
a² = p² ; a = p
b² = 16 = 4² ; b = 4
2ab = 2*p*4 = 8p
p² - 8p + 16 = (p - 4)²
ii) a² + 2ab + b² = (a +b)²
a² = 121x² = (11x)² ; b² = 4y² = (2y)²
2ab = 2 * 11x * 2y = 44xy
121x² + 44xy + 4y² = (11x + 2y)²
Answer:
a) x1 = 6 and x2 = -2
b) -2
Step-by-step explanation:
a)
To find the roots of the quadratic equation, we can use the Bhaskara's formula:
Delta = b^2 - 4ac
Delta = (-4)^2 - 4*1*(-12) = 64
sqrt(Delta) = 8
x1 = (-b + sqrt(Delta)) / 2a
x1 = (4 + 8) / 2
x1 = 6
x2 = (-b - sqrt(Delta)) / 2a
x2 = (4 - 8) / 2
x2 = -2
b)
The roots are 6 and -2, so the smaller root is -2
π I don’t really know the answer I’m really sorry
