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
Answer:
-2, -5
Step-by-step explanation:
h2 + 7h + 10 = 0
h^2 +7h + 10 = 0
h^2 +2h +5h + 10 = 0
h(h+2) + 5(h+2) = 0
h+2 = 0 or h+5 = 0
h = -2 or h = -5
The formula for compound interest
A = P( 1 + r/n) ^ (nt)
A is the amount in the account at the end
P is the principal balance or the amount initially invested
r is the annual interest rate in decimal form
n is the number of times it is coupounded per year
t is the number of years
A = 1800 ( 1+ .0375/1) ^ (1*6)
A = 1800 ( 1.0375)^6
A = 2244.92138
Rounding to the nearest cent
A = 2244.92
Answer: 12.50$
Explanation: 5.25$ (per pound) and he wants 3 pounds so we multiply 5.25x3 which equals 15.75 and then we subtract 3.25 (because of the coupon) which makes your final answer 12.50$
Answer: 55.1°F
Step-by-step explanation:
We are informed that there is a 1.6°F drop in temperature for every thousand feet that an airplane climbs into the sky, if the plane reaches an altitude of 2,000ft, the drop in temperature will be:
= 1.6° × 2
= 3.2°F
If the temperature on the ground is 58.3°F, then the temperature when the plane reaches an altitude of 2,000 ft will be:
= 58.3°F - 3.2°F
= 55.1°F
