There's a theorem that states:
"<span>If a quadrilateral is a parallelogram, </span>it has<span> 2 sets of opposite sides congruent.</span><span>"
</span>
Hope this helps ;)
Answer:
y" = csc(x)[9cot²(x) - csc²(x)]
Step-by-step explanation:
Step 1: Define
y = 9csc(x)
Step 2: Find 1st derivative
y' = -9csc(x)cot(x)
Step 3: Find 2nd derivative
y" = 9csc(x)cot(x)cot(x) + -csc(x)csc²(x)
y" = 9csc(x)cot²(x) - csc³(x)
y" = csc(x)[9cot²(x) - csc²(x)]
Set up the equation so m<HCB + m<DCH = m<DCB
2x + 100 + 60 = x + 170
2x+ 160 = x +170
-x -x
x + 160 = 170
- 160 -160
x = 10
Answer:
Step-by-step explanation:
hello :
Part A : x+6y =6 means : 6y = - x+6
so : y = (-1/6)x+1 an equation for the line (D)
y = (1/3)x -2 is the line (D')
PartB : solution of the system : y = (-1/6)x+1 ....(1) color red
y = (1/3)x -2 ....(2) color bleu
is the intersection point : (6 ; 0)
PartC : Algebraically by (1) and (2) : (-1/6)x+1 = (1/3)x -2
(-1/6)x - (1/3)x = -2-1
(-x-2x)/6= -3
-3x = -18
so : x = 6 put this value in (1) or (2) : y = (-1/6)(6)+1 =0 the solution is : (6 ;0)
A. let the distance be d
so we have Δd/Δt=8.8min
Δd= 8.8Δt
d=∫8.8Δt-------(1)
b. d should be positive because the diver is ascending at a positive rate relative to depth
c. d=rt
t=d/r = 32.12/8.8 min
t= 3.65min