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3 years ago

Factor completely:

Mathematics

1 answer:

padilas [110]3 years ago

8 0

Answer:

d) 2x²(6x³ + 3x + 4)

Step-by-step explanation:

In order to factor the given polynomial, you need to find the greatest common factor of all three terms: $12x^{5}+6x^{3}+8x^{2}$

The greatest common factor is '2' and the variable is x². If we factor out 2x² from each term, we get:

2x²(6x³ + 3x + 4)

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3 years ago

Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dy

mixas84 [53]

Incomplete question as we have not told what to find.So I have assumed to find average CPI for each program.So complete question is here

Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dynamic instruction count of 1.0E9 and has an execution time of 1.1 s, while compiler B results in a dynamic instruction count of 1.2E9 and an execution time of 1.5 s.

Find the average CPI for each program given that the processor has a clock cycle time of 1 ns.

Answer:

$CPI_{A}=1$

$CPI_{B}=1.25$

Step-by-step explanation:

As we know CPU time given as

$CPU_{time}=instructions*CPI*Cycle_{time} \\So\\CPI=\frac{CPU_{time}}{instructions*Cycle_{time}}\\ Where\\Cycle_{time}=1*10^{-9}s\\ And\\executionTime=CPUtime$

For Compiler A

$CPI_{A}=\frac{CPU_{timeA} }{instruction_{A}*Cycle_{time} }\\CPI_{A}=\frac{1s}{10^{9}*1.0*10^{-9}s }\\CPI_{A}=1$

For Compiler B

$CPI_{B}=\frac{CPU_{timeB} }{instruction_{B}*Cycle_{timeB} }\\CPI_{B}=\frac{1.5s}{(1.2*10^{9})*(1.0*10^{-9}s) }\\CPI_{B}=1.25$

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4 years ago