<span>y = sqrt(25-x^2) at point (3,4)
The derivative gives us the slope at 3 to be:
-2x
------------ at x=3: -3/4
2sqrt(25-x^2)
</span><span>so we have a vector that is parallel to the slope of the tangent line is: <4,-3>
</span>
<span>the mag = 5 so; unit tangent = <4/5 , -3/5>
</span>
<span>since perpendicular lines have a -1 product between slopes we get the normal to be...
<3/5,4/5>
</span>
<span>It is <4,-3> because it is rise over run. Rise is y component of vector and run is x component of vector.</span>
Answer:
LOM= 130º
Step-by-step explanation:
Combine the two equations and set it to equal 180º(LON), 180º=12x+120º
next subtract 120º from both sides, 60º=12x. Now divide both sides by 12, 5=x. Now you can plug 5 back into the original equation to calculate MON, 8x5+90º
The first three you should check off because the man lost less so the last few questions would be wrong
Answer: the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
Step-by-step explanation:
Given data;
lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min
hence, height = 1 / ( 52.0 - 50.0) = 1 / 2
now the probability that the class length is between 50.8 and 51 min = ?
P( 50.8 < X < 51 ) = base × height
= ( 51 - 50.8) × 1/2
= 0.2 × 0.5
= 0.1 ≈ 10%
therefore the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
I dont have time to do all of these right now my apolgies but Ill explain how to do it.
The 3 angles of a triangle add up to 180.
So let me show you how to do 1.
.
.
.
1. 58 degrees
92, 30, ?
If we add together the first two angles,
92+30= 122
Knowing the total addition of all three angles would ne 180, we can subtract 122 from it to get the third angle.
180-122= 58
So we now know the last angle is 58 degrees because
92+30+58=180
