Answer:
i. ΔAXC ~ ΔCXB
ii. ΔBCX Is-congruent-to ΔACX
Step-by-step explanation:
From the given ΔABC, CX is the altitude of ΔABC; and also an angle bisector of <ACB.
So that:
m<AXC = m<BXC (right angle property)
m<ACX = m<BCX (congruent property)
m<ACX + m<AXC + m<CAX = 
(sum of angles in a triangle)
m<BCX + m<BXC + m<CBX = (sum of angles in a triangle)
Therefore, from the figure it can be deduced that;
i. ΔAXC ~ ΔCXB (Angle-Angle-Side, AAS, property)
ii. ΔBCX Is-congruent-to ΔACX (Angle-Angle-Side, AAS, property)
46/45. do u know how 2 do it? hope u do
Answer:
h(-5) = 34
Step-by-step explanation:
h(x) = -5x + 9
h(-5) = -5(-5) + 9 = 25 + 9 = 34
Answer:
1. y' = 3x² / 4y²
2. y'' = 3x/8y⁵[(4y³ – 3x³)]
Step-by-step explanation:
From the question given above, the following data were obtained:
3x³ – 4y³ = 4
y' =?
y'' =?
1. Determination of y'
To obtain y', we simply defferentiate the expression ones. This can be obtained as follow:
3x³ – 4y³ = 4
Differentiate
9x² – 12y²dy/dx = 0
Rearrange
12y²dy/dx = 9x²
Divide both side by 12y²
dy/dx = 9x² / 12y²
dy/dx = 3x² / 4y²
y' = 3x² / 4y²
2. Determination of y''
To obtain y'', we simply defferentiate above expression i.e y' = 3x² / 4y². This can be obtained as follow:
3x² / 4y²
Let:
u = 3x²
v = 4y²
Find u' and v'
u' = 6x
v' = 8ydy/dx
Applying quotient rule
y'' = [vu' – uv'] / v²
y'' = [4y²(6x) – 3x²(8ydy/dx)] / (4y²)²
y'' = [24xy² – 24x²ydy/dx] / 16y⁴
Recall:
dy/dx = 3x² / 4y²
y'' = [24xy² – 24x²y (3x² / 4y² )] / 16y⁴
y'' = [24xy² – 18x⁴/y] / 16y⁴
y'' = 1/16y⁴[24xy² – 18x⁴/y]
y'' = 1/16y⁴[(24xy³ – 18x⁴)/y]
y'' = 1/16y⁵[(24xy³ – 18x⁴)]
y'' = 6x/16y⁵[(4y³ – 3x³)]
y'' = 3x/8y⁵[(4y³ – 3x³)]
Answer:
-8 and 9
Step-by-step explanation:
-8 x 9 = -72
and -8 + 9 = 1
