Answer:
∠EFG = 48°
Step-by-step explanation:
As FH bisects ∠EFG , ∠EFH = ∠HFG .
We know that ∠EFH = (-5x + 89)° . So ∠HFG = ∠EFH = (-5x + 89)°
Also, ∠HFG + ∠EFH = ∠EFG
=> 2(-5x + 89)° = (61 - x)°
=> -10x + 178 = 61 - x
=> 10x - x = 178 - 61
=> 9x = 117
=> x = 117 / 9 = 13
Putting the value of 'x' in ∠EFG gives :-
(61 - x)° = (61 - 13)° = 48°
Let's call bicycles 'b' and unicycles 'u'
Note that bicycles have 2 tires, and unicycles have 1 tire.
We can make two equations:
b = u + 8
2b + 1u = 46
Solving the first equation for u, we get:
u = b - 8
Plug that equation into the second, and we get:
2b + (b - 8) = 46
Subtract 2b on both sides.
1(b - 8) = 46 - 2b
Basically, I used b and your question used n.
The correct answer is: D. 1(n - 8) = 46 - 2n.
The length of the KN is 4.4
Step-by-step explanation:
We know from Pythagoras theorem
In a right angle ΔLMN
Base² + perpendicular² = hypotenuse
²
From the properties of triangle we also know that altitudes are ⊥ on the sides they fall.
Hence ∠LKM = ∠NKM = 90
°
Given values-
LM=12
LK=10
Let KN be “s”
⇒LN= LK + KN
⇒LN= 10+x eq 1
Coming to the Δ LKM
⇒LK²+MK²= LM²
⇒MK²= 12²-10²
⇒MK²= 44 eq 2
Now in Δ MKN
⇒MK²+ KN²= MN²
⇒44+s²= MN² eq 3
In Δ LMN
⇒LM²+MN²= LN²
Using the values of MN² and LN² from the previous equations
⇒12² + 44+s²= (10+s)
²
⇒144+44+s²= 100+s²+20s
⇒188+s²= 100+s²+20s cancelling the common term “s²”
⇒20s= 188-100
∴ s= 4.4
Hence the value of KN is 4.4
Step-by-step explanation:
(1\2)^4 x 1\16 x 1\8 x 1\4 [1/2^4=16]
1/16 x 1/512
=>1/2048
Answer:
141°
Step-by-step explanation:
It's an octagon so the Sum of interior angles is: (8-2)×180=1080
5x+133+16+126+162-12+30=5x+455=1080
5x=625
x=125
x+16=141
