The length of the measure of YZ is 80cm
<h3>Perpendicular lines</h3>
Perpendicular lines are lines that are at angle 90 degrees to each other. From the given diagram;
WX = 82cm
Required
Length. of YZ
Using the pythagoras theorem
YV = VZ = √41²-9²
YV = VZ = √1600
YV = VZ = 40
Since YZ = YV + VZ
YZ = 40cm + 40cm
YZ= 80cm
Hence the length of the measure of YZ is 80cm
Learn more on perpendicular lines here: brainly.com/question/1202004
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Answer:
AB = 142
Step-by-step explanation:
If M is the midpoint, then AM = MB
6x+5 = 7x -6
Subtract 6x from each side
6x-6x+5 = 7x-6x -6
5 = x-6
Add 6 to each side
5+6 = x-6+6
11=x
We want to find AB
AB = 6x+5+7x-6
AB = 13x -1
= 13(11) -1
=143-1
= 142
Answer:
Step 2 has the error
1. The whole number (20) was left out
2. Adding different signs, we subtract and retain the sign from the larger number
Correction:
Step 2: 20 - 17x = 34x + 60
Step 3: -17x -34x = 60 - 20
Step 4: -51x = 40
Step 5: x = 40/-51
3 terms
https://youtu.be/0sq2PMQ_Nak
use that video to help you.
Answer:
x = 1 - 5t
y = t
z = 1 - 5t
Step-by-step explanation:
For the equation of a line, we need a point and a direction vector. We are given a point (1, 0, 1).
Since the line is suppose to be a tangent to the given curve at the point (1, 0, 1), we need to find a tangent vector for which the curve passes through that point.
We have x = e^(-5t)cos5t
at t = 1, x = e^(-5)cos5
at t = 0, x = 1
y = e^(-5t)sin5t
at t = 1, y = e^(-5)sin5
at t = 0, y = 0
z = e^(-5t)
at t = 1, z = e^(-5)
at t = 0, z = 1
Clearly, the only parameter value for which the curve passes through the point (1, 0, 1) is t = 0.
In vector notation, the curve
r(t) = xi + yj + zk
= e^(-5t)cos5t i + e^(-5t)sin5t j + e^(-5t) k
r'(t) = [-5e^(-5t)cos5t - e^(-5t)sin5t] i +[e^(-5t)cos5t - 5e^(-5t)sin5t] j - 5e^(-5t) k
r'(0) = -5i + j - 5k
is a vector tangent at the point.
We get the parametric equation from this.
x = x(0) + tx'(0)
= 1 - 5t
y = y(0) + ty'(0)
= t
z = z(0) + tz'(0)
= 1 - 5t
