The equation between the number of days remaining and number of hours is given as y= -5x+50.
<h3>How to illustrate the equation?</h3>
Let y be the number of days remaining and x be the no.of hours.
From the graph the slope of the line is given by m.
m=(y2 - y1)/(x2 - x1)
m=(30 - 40)/(4 - 2)
m= -10/2 = -5.
The above slope is in between points 2 and 4. As the line has a constant rate of growth the slope is the same between any points of the line.
Therefore the equation between the no.of days remaining and no.of hours is given by
y=mx +c.
y=-5x+c
Now to find the value of c. This can be determined by substituting the x and y values at a particular instant. For x=0,y=50. Therefore by these values, we get,
50= -5(0)+c
c=50.
Therefore the relation between the number of days remaining and number of hours is given as y= -5x+50.
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Answer:
volume of cylinder = 1232cm³
Step-by-step explanation:
from the question 
volume of cylinder =?
height of the cylinder = 8cm
Base diameter of the cylinder = 14cm
 to solve this problem we need to know the formula to calculate the volume of a cylinder
volume of cylinder = πr²h 
                    or
volume of cylinder = π(d/2)²h........................... if we are given diameter and not radius.
D= 14cm 
d = 14/2
d = 7cm
now,
volume of cylinder = π × (7)² × 8
volume of cylinder = 22/7 ×49 ×8
volume of cylinder = 22 × 7 × 8
volume of cylinder = 1232cm³
therefor the volume of the cylinder is equals to 1232cm³
 
        
             
        
        
        
Answer:
5/4k^2
Step-by-step explanation:
P=5\dfrac{k}{6}\times \dfrac{3}{2k^3}.
We will be using the following property of exponents:
\dfrac{a^x}{a^y}=a^{x-y}.
We have
P\\\\\\=5\dfrac{k}{6}\times\dfrac{3}{2k^3}\\\\\\=\dfrac{5}{6}\times\dfrac{3}{2}k^{1-3}\\\\\\=\dfrac{5}{4}k^{-2}=\dfrac{5}{4k^2}.
Thus, the required product is \dfrac{5}{4k^2}.
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Answer:
it might be 3 one according to me