Given:
The sum of 8 and B is greater than 22.
To find:
The inequality for the given statement and its solution.
Solution:
We know that, sum of two number is the addition of two numbers.
Sum of 8 and B = 8+B
It is given that, the sum of 8 and B is greater than 22.

Subtracting 8 from both sides, we get


Therefore, the required inequality for the given statement is
and the solution is
.
Answer:
C.h=infinitely many solution
Answer:
0.3*10+10/5
PEMDAS (left to right)
3+2
5
Step-by-step explanation:
a.
has an average value on [5, 11] of

b. The mean value theorem guarantees the existence of
such that
. This happens for

Answer:
<em>The mass of the steel ball is 4,235.9 gr</em>
Step-by-step explanation:
<u>Density</u>
The density ρ of a substance is a measure of its mass per unit volume:

If the density and the volume are given, the mass can be calculated by solving the above formula for m:

We know the density of pure steel ρ=8.09 gr/cm3 and the diameter of a solid steel ball d=10 cm.
We need to calculate the volume of the sphere:
The volume of a sphere of radius r is given by:

The radius is half the diameter: r= 10/2 = 5 cm. Thus:

Calculating:

The mass is:

m=4,235.9 gr
The mass of the steel ball is 4,235.9 gr