Sides:6.7,9.3,9.5
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The equation has one extraneous solution which is n ≈ 2.38450287.
Given that,
The equation;

We have to find,
How many extraneous solutions does the equation?
According to the question,
An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.
To solve the equation cross multiplication process is applied following all the steps given below.

The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y
with 0 and solve for x. The graph of the equation is attached.
n ≈ 2.38450287
Hence, The equation has one extraneous solution which is n ≈ 2.38450287
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Answer:
Radius of sphere is 3 units.
Step-by-step explanation:
Volume of sphere is given by 
surface area of sphere is given by 
where r is the radius of the sphere.
Given that
The number of cubic units in the volume of a sphere is equal to the number of square units in the surface area of the sphere.
we equate formula of Volume of sphere and surface area of sphere
assuming r as the radius.
thus,

Thus, radius of sphere is 3 units.
Answer:
Yes.
Step-by-step explanation:
Just like normal algebra, you factor our the common factor, in this case, 5.
Thus,

Answer:
b+10g-x
Step-by-step explanation:
3x-4x=x
New equation becomes:
b+10g-x