The height of a box is 3 inches. The length is three inches more than the width. Find the width if the volume is 390. Select one
: A. 13 in. B. 130 in. C. 10 in. D. 3 in.
1 answer:
Let
x---------> the length side of the box
y---------> the width side of the box
we know that
h=3 in
V=390 in³
volume of the box=x*y*3-------> 390=x*y*3-----> x*y=130------> equation 1
x=3+y------> equation 2
substitute 2 in 1
[3+y]*y=130-----> y²+3y=130------> y²+3y-130=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solution is
y=10 in
x=3+y-----> x=3+10-----> x=13 in
the answer is
the width is 10 in
x---------> the length side of the box
y---------> the width side of the box
we know that
h=3 in
V=390 in³
volume of the box=x*y*3-------> 390=x*y*3-----> x*y=130------> equation 1
x=3+y------> equation 2
substitute 2 in 1
[3+y]*y=130-----> y²+3y=130------> y²+3y-130=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solution is
y=10 in
x=3+y-----> x=3+10-----> x=13 in
the answer is
the width is 10 in
You might be interested in
Answer: (3, -2)
Step-by-step explanation:
The solution to the system is where the graphs intersect.
If a function is defined as
where both are continuous functions, then
is also continuous where defined, i.e. where
So, in your case, this function is continous everywhere, except where
To solve this equation, we can use the formula
It means that, if the leading terms is 1, then the x coefficient is the opposite of the sum of the roots, and the constant term is the product of the roots.
So, we're looking for two terms whose sum is 7, and whose product is 12. These numbers are easily found to be 3 and 4.
So, this function is continuous for every real number different than 3 or 4.