The answer is C
9 - 4 = 5
5 is equal to 5
13 - 4 = 9
9 is greater than 5
The symbol means greater than or equal to
Answer:
y=-4/3x-9
=
−
4
3
x
−
9
Step-by-step explanation:
First we need to get the slope of the perpendicular line
The slope of a perpendicular line is equal to the negative inverse of the slope of the given line
y
=
m
x
+
b
y
=
3
4
x
−
2
⇒
m
=
3
4
m
'
=
−
1
m
⇒
m
'
=
−
1
3
4
⇒
m
'
=
−
4
3
Now that we have the slope, we need to find the y-intercept.
To find the y-intercept, we need to plug-in values of
x
and
y
that the line passes through
y
'
=
m
'
x
'
+
b
⇒
y
'
=
−
4
3
x
'
+
b
⇒
7
=
−
4
3
(
−
12
)
+
b
⇒
7
=
16
+
b
⇒
b
=
−
9
Therefore, the equation of the line is
y
=
−
4
3
x
−
9
Answer:
7.3 answer
Step-by-step explanation:
x=1.7
now
3x+2.2
3×1.7+2.2
7.3
Answer:
It will take <u><em>80 days</em></u> for the bull calf to reach a weight of 500 kilograms.
Step-by-step explanation:
Given:
The weight of a bull calf is 388 kilograms.
Now, to find the weight of bull calf of how long it will take to reach a weight of 500 kilograms, if it’s weight increases at a rate of 1 2/5 kilograms per day.
Required weight which to be increased = 500 - 388 = 112 kilograms.
Rate of weight increase = 
=
Thus, the time required = 
=
=
<em>The time required = 80 days</em>.
Therefore, it will take 80 days for the bull calf to reach a weight of 500 kilograms.
Answer:
See explanation
Step-by-step explanation:
Jane is given a $100 gift to start and saves $35 a month from her allowance.
- After 1 month, Jane has saved

- After 2 months, Jane has saved

- After three months, Jane has saved

- and so on
In general, after x months Jane has saved

This means that it makes sense to represent the relationship between the amount saved and the number of months with one constant rate (in this case the constant rate is 35). It makes sense because the amount of money increases by $35 each month. Since the amount of increase is constant, we get constant rate. Also the initial amount is known ($100), so there is a possibility to write the equation of linear function representing this situation.