The slope is -30 because when u calculate the rise over run it gets to -30
Answer:
90t-21t=69t
Step-by-step explanation:
4:00pm to 6:00 am is 14 hours. 14 x 1.5 is in fact 21 which means the temperature dropped 21 degrees in those 14 hours.
Hey There!
Here is your answer:
First write the equation down:
-3 4/5 ÷ -3 1/6
Then put the mixed fractions into whole numbers:
-3 4/5= 5×-3=-15+4=-11= -11/5
&
-3 1/6= 6×-3=-18+1=-17= -17/6
Now right the new problem:
-11/5×6/-17
Now multiply:
-11×6=-66
5×-17=-85
Which means -66/-85 is your answer!
Hope this helps!
Answer:
Dimensions: 
Perimiter: 
Minimum perimeter: [16,16]
Step-by-step explanation:
This is a problem of optimization with constraints.
We can define the rectangle with two sides of size "a" and two sides of size "b".
The area of the rectangle can be defined then as:

This is the constraint.
To simplify and as we have only one constraint and two variables, we can express a in function of b as:

The function we want to optimize is the diameter.
We can express the diameter as:

To optimize we can derive the function and equal to zero.

The minimum perimiter happens when both sides are of size 16 (a square).