surface area (S) of a right rectangular solid is:
S = 2*L*W + 2*L*H + 2*W*H (equation 1)
where:
L = length
W = width
H = height
-----
you have:
L = 7
W = a
H = 4
-----
formula becomes:
S = 2*7*a + 2*7*4 + 2*a*4
simplify:
S = 14*a + 56 + 8*a
combine like terms:
S = 22*a + 56
-----
answer is:
S = 22*a + 56 (equation 2)
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to prove, substitute any value for a in equation 2:
let a = 15
S = 22*a + 56 (equation 2)
S = 22*15 + 56
S = 330 + 56
S = 386
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since a = 15, then W = 15 because W = a
go back to equation 1 and substitute 15 for W:
S = 2*L*W + 2*L*H + 2*W*H (equation 1)
where:
L = length
W = width
H = height
-----
you have:
L = 7
W = 15
H = 4
-----
equation 1 becomes:
S = 2*7*15 + 2*7*4 + 2*15*4
perform indicated operations:
S = 210 + 56 + 120
S = 386
-----
surface area is the same using both equations so:
equations are good.
formula for surface area of right rectangle in terms of a is:
S = 22*a + 56
-----
Two out of five Would be my answer
Answer: y = 4x + 43
Step-by-step explanation: In the photo
Answer:
6% chance
Step-by-step explanation:
12+19+14+15 = 60 60 divided by 100 = 0.6 = 6%
The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.
Domain is the values of x in the function represented by y=f(x), for which y exists.
THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".
Now we assume the 
and 
So here since g(x) is a polynomial function so it exists for all real x.
<em> </em>does not exists when 
, so the domain of f(x) is given by all real x except 6.
Now,
So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.
But domain of both f(x) and g(x) consists of the value x=4.
Hence the statement is not TRUE universarily.
Thus the given statement about the composition of function is FALSE.
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