Part I)
The module of vector AB is given by:
lABl = root ((- 3) ^ 2 + (4) ^ 2)
lABl = root (9 + 16)
lABl = root (25)
lABl = 5
Part (ii)
The module of the EF vector is given by:
lEFl = root ((5) ^ 2 + (e) ^ 2)
We have to:
lEFl = 3lABl
Thus:
root ((5) ^ 2 + (e) ^ 2) = 3 * (5)
root ((5) ^ 2 + (e) ^ 2) = 15
Clearing e have:
(5) ^ 2 + (e) ^ 2 = 15 ^ 2
(e) ^ 2 = 15 ^ 2 - 5 ^ 2
e = root (200)
e = root (2 * 100)
e = 10 * root (2)
as AB || DE then
angle BAE = angle AED
so angle AED = 70
as AD = AE then
angle AED = angle ADE = 70
now in a triangle
angle DAE + Angle ADE + angle AED = 180
angle DAE + 70 + 70 = 180
angle DAE = 180 - 140 = 40
Answer:
Step-by-step explanation:
Due to the fact that the degree of the polynomial is odd, we know one end must go up and the other down. Because the leading coefficient is even, we know that as x approaches negative infinity, y approaches negative infinity as well. As x approaches positive infinity, y approaches positive infinity.
Answer: x=12
Step-by-step explanation:
we know that sum of a triangle is 180 degrees
one angle is 90 degree
the sum will be 90+(2x+1)+(5x+5)=180
substract 90 from both sides
2x+1+5x+5=180-90
7x+6=90
7x=90-6
divide by 7
x=12
Answer:
the answer is C
Step-by-step explanation:
