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3 years ago

15

If a simple pipelined processor is super-pipelined by a factor of 3 (the ALU takes 3 cycles instead of one for the smallest oper

ation), back-to-back dependent instructions require ______ stall cycles to be inserted even with perfect bypassing.

Computers and Technology

1 answer:

GrogVix [38]3 years ago

4 0

Answer: 2

Explanation:

If it is super pipelined with ALU taking 3 cycles the back to back dependent instructions require to insert 2 stall cycles as the control unit can stall the execution of the pipeline as as the ALU takes time in execution using 3 cycles. This is done so in order to avoid a hazard in the pipeline.

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Which transformation(s) can be used to map ARST onto

xenn [34]

Answer:

The last line i.e "rotation, then translation " is the correct answer .

Explanation:

Missing information :

In this question some information is missing The triangles are congruent by the SSS and HL. This line is missing in the given question .

The triangles that are the exactly similar sides and the angles.is known as congruent triangles .In the congruent triangle One's three sides are exactly similar in way of measuring to someone else's three sides also each of the three angles has almost the same angle as another.

In the congruent triangles firstly the reflection will applied then after the translation process is applied in the SSS and HL.
The other options are incorrect because they are not used for the mapping of RST into AWWX that's why these are incorrect option .

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2 years ago

Find the inverse function of f(x)= 1+squareroot of 1+2x

Svetllana [295]

Answer:

Therefore the inverse function of $f(x)=1+\sqrt{1+2x}$ is $\frac{x^2-2x}{2}$

Explanation:

We need to find the inverse of function $f(x)=1+\sqrt{1+2x}$

Function Inverse definition :

$\mathrm{If\:a\:function\:f\left(x\right)\:is\:mapping\:x\:to\:y,\:then\:the\:inverse\:functionof\:f\left(x\right)\:maps\:y\:back\:to\:x.}$

$y=1+\sqrt{1+2x}$ $\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y$

$x=1+\sqrt{1+2y}$

$\mathrm{Solve}\:x=1+\sqrt{1+2y}\:\mathrm{for}\:y$

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$1+\sqrt{1+2y}-1=x-1$

Simplify

$\sqrt{1+2y}=x-1$

$\mathrm{Square\:both\:sides}$

$\left(\sqrt{1+2y}\right)^2=\left(x-1\right)^2$

$\mathrm{Expand\:}\left(\sqrt{1+2y}\right)^2:\quad 1+2y$

$\mathrm{Expand\:}\left(x-1\right)^2:\quad x^2-2x+1$

$1+2y=x^2-2x+1$

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$1+2y-1=x^2-2x+1-1$

$\mathrm{Simplify}$

$2y=x^2-2x$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\frac{2y}{2}=\frac{x^2}{2}-\frac{2x}{2}$

$\mathrm{Simplify}$

$y=\frac{x^2-2x}{2}$

Therefore the inverse function of $f(x)=1+\sqrt{1+2x}$ is $\frac{x^2-2x}{2}$

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Answer:

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