Please help me with this homework
2 answers:
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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:
View image
Step-by-step explanation:
You treat inequality questions just like normal equation with an equal sign. You only have to worry about the inequality when you do the shading part.
Your first goal is to solve for y, which I did at the top.
And remember that when you divide/multiply by a negative number you have to flip the inequality sign.
So the first equation is already in term of y. (y < 2x + 4)
So you only have to solve for y for -3x - 2y ≥ 6.
I solved that and got y ≤ -3/2x - 3.
Now you have 2 equations. y < 2x + 4 and y ≤ -3/2x - 3
First you gotta treat it like a normal equation and graph them.
So i graphed y = 2x + 4 and y = -3/2x - 3. I assume you already know how do do that.
btw. < and > have dotted lines, while ≤ and ≥ have solid lines.
Now for the shading, you gotta look at the inequality sign.
y < 2x + 4 has a less than sign, so you have to shade everything underneath the line you graphed.
y ≤ -3/2x - 3 is also less than, so you have to shade everything underneath that line as well.
So in the future, you you get a > or ≥ then you have to shade above the graph.
