Length of x is 98.2 m
<u>Step-by-step explanation:</u>
Step 1:
Use the trigonometric ratio tan 27° to find the common side of both the right angled triangles.
tan 27° = opposite side/adjacent side = opposite side/9
∴ Opposite side = 9 tan 27° = 9 × - 3.27 = -29.46 m
Step 2:
Use this side and trigonometric ratio cosine to find the value of x.
cos 49° = adjacent side/x = -29.46/x
∴ x = -29.46/cos 49° = -29.46/0.30
= 98.2 m (negative value neglected)
Answer:
B
Step-by-step explanation:
ABD (half of the rectangle) is a right-angled triangle.
so, Pythagoras applies.
c² = a² + b²
c is the Hypotenuse (the baseline opposite of the 90 degree angle) and in our case the line BD.
a and b are the 2 sides enclosing the 90 degree angle. in our case here the lines AD and AB.
so,
BD² = AD² + AB² = (6-1)² + (5-2)² = 5² + 3² = 25 + 9 = 34
BD = sqrt(34)
Answer:
3x+y=8 and 3x+y=9
Step-by-step explanation:
Assume these two equations:
a1x+b1y+c1=0
a2x+b2y+c2=0
If (a1/a2) = (b1/b2) ≠ (c1/c2) than those two would have no solution.
In these case, take a look at the last option:
Re-write 'em as a standard form:
So (a1/a2) = (b1/b2) ≠ (c1/c2) is true here:
And they do not have any collision point
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
