Considering the relation built the presence of point M on line LN, the numerical length of LN is of 9 units.
<h3>What is the relation from the presence of point M on the line LN?</h3>
Point M splits line LN into two parts, LM and MN, hence the total length is given by:
LN = LM + MN.
From the given data, we have that:
Hence we first solve for x.
LN = LM + MN.
2x - 5 = 3 + x - 1
x = 7.
Hence the total length is:
LN = 2x - 5 = 2 x 7 - 5 = 14 - 5 = 9 units.
More can be learned about relations and lines at brainly.com/question/2306122
#SPJ1
If Jordan has $55 and she earns $67 by doing chores, you can calculate how much money does Jordan have now using the following step:
$55 + $67 = $122
Result: Jordan has $122.
Answer:
Step-by-step explanation:
is the given line 3x - 4y = -17?
then,
- 4y = -3x -17
y=(3/4)x+17/4
it is parallel, so slope of our line is 3/4 or 0.75
y=0.75x+b
next, lets plug in the given point to get the y-int
2=0.75*(-3)+b
2=-2.25+b
b-2.25=2
b=4.25
the equation is y=0.75x+4.25
does this clear anything, or am I off topic? please tell me I will help you
<h2>T
he car is about 6.6 years old.</h2>
Step-by-step explanation:
Given : An equation for the depreciation of a car is given by
, where y = current value of the car, A = original cost, r = rate of depreciation, and t = time, in years. The value of a car is half what it originally cost. The rate of depreciation is 10%.
To find : Approximately how old is the car?
Solution :
The value of a car is half what it originally cost i.e. 
The rate of depreciation is 10% i.e. r=10%=0.1
Substitute in the equation, 


Taking log both side,




Therefore, the car is about 6.6 years old.
The only way I could ever do that is 3