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D-Link Corporation (: 友訊科技) is a Taiwanese multinational
manufacturing corporation headquartered in , Taiwan. It was founded in March 1986 in Taipei as Datex Systems Inc.
D-Link Corporation changed its name from Datex Systems Inc. in 1994, when it went public and when it became the first networking company on the . It is now publicly traded on the TSEC and NSE stock exchanges.
In 2007, it was the leading networking company in the small to medium business (SMB) segment worldwide with 21.9% market share. In March 2008, it became the market leader in Wi-Fi product shipments worldwide, with 33% of the total market. In 2007, the company was featured in the "Info Tech 100", a listing of the world's best IT companies. It was also ranked as the 9th best IT company in the world for shareholder returns by .
D-Link's products are geared towards the networking and communications market. Its business products include switches, surveillance network cameras, firewalls, iSCSI SANs and business wireless, while consumer products cover consumer wireless devices, broadband devices, and the Digital Home devices (which include media players, storage, and surveillance camera/NVR).
AC5300 MU-MIMO Ultra Wi-Fi Router
AC1900 Wi-Fi USB 3.0 Adapter
mydlink Home Smart Plug
mydlink Home Monitor 360
In January 2010, it was reported that
vulnerabilities had been found on some D-Link routers. D-Link was also criticized for their response which was deemed confusing as to which models were affected and downplayed the seriousness of the risk.
In January 2013, version v1.13 for the DIR-100 revA was reported to include a backdoor in the firmware. By passing a specific
in an HTTP request to the router, normal authentication is bypassed. It was reported that this backdoor had been present for some time.
Computerworld reported in January 2015 that ZynOS, a firmware used by some D-Link routers (as well as , , and others), are vulnerable to
by an unauthenticated remote attacker, specifically when remote management is enabled.
Later in 2015, it was reported that D-Link leaked the private keys used to sign firmware updates for the DCS-5020L security camera and a variety of other D-Link products. The key expired in September 2015, but had been published online for seven months.
Also in 2015, D-Link was criticized for more HNAP vulnerabilities, and worse, introducing new vulnerabilities in their "fixed" firmware updates.
On 5 January 2017, the
sued D-Link for failing to take reasonable steps to secure their routers and IP cameras. As D-Link marketing was misleading customers into believing their products were secure. The complaint also says security gaps could allow hackers to watch and record people on their D-Link cameras without their knowledge, target them for theft, or record private conversations. D-Link has denied these accusations and has enlisted Cause of Action Institute to file a motion against the FTC for their "baseless" charges.
In 2006, D-Link was accused of , when it was found that its routers were sending time requests to a small
in Denmark, incurring thousands of dollars of costs to its operator. D-Link initially refused to accept responsibility. Later, D-link products were found also to be abusing other time servers, including some operated by the US military and .
On 6 September 2006, the gpl-violations.org project prevailed in court litigation against D-Link Germany GmbH regarding D-Link's alleged inappropriate and copyright infringing use of parts of the .
. "." Retrieved 11 July 2012.
Compiled from In-Stat Q1 2007 Wireless LAN Equipment Market Share Report
In-Stat Q4/07 WLAN Market Share Report
BusinessWeek Magazine, "Info Tech 100" – Issue 2 July 2007
. 18 January 2010. Archived from
on 26 December 2013.
Yegulalp, Serdar. . InfoWorld.
Constantin, Lucian. . Computerworld.
. Ars Technica.
. Cause of Action Institute. .
Leyden, John. . The Register.
Ward, Mark. . BBC.
7 October 2014 at the .
Media related to
at Wikimedia Commons
: Hidden categories:Evaluation: Evaluating Expressions, Polynomials, and Functions
Evaluation:
Evaluating
&&&&Expressions, Polynomials, and
Functions (page
Sections: Evaluating Expressions and
Polynomials,
&Evaluation&
mostly means &simplifying an expression down to a single numerical
value&. Sometimes you will be given a numerical expression, where
all you ha that is more of an
kind of question. In this lesson, I'll concentrate on the &plug and
chug& aspect of evaluation: plugging in values for variables, and
&chugging& my way to the simplified answer.
Usually the
only hard part in evaluation is in keeping track of the minus signs. I
would strongly recommend that you use parentheses liberally, especially
when you're just getting started.
Evaluate a2b for a = &2, b = 3, c = &4,
and d = 4.
To find my answer, I
just plug in the given values, being careful to use parentheses, particularly
around the minus signs:
&&Copyright
& Elizabeth Stapel
All Rights Reserved
= (4)(3) = 12
Evaluate a & cd for a = &2, b = 3, c = &4,
and d = 4.
(&2) & (&4)(4)
= &2 & (&16) = &2 + 16 = 16 & 2 = 14
Evaluate (b + d)2 for a = &2, b = 3, c = &4,
and d = 4.
I must take care not
to try to &distribute& the exponent through the parentheses.
Exponents do NOT distribute over addition! I should never try to say
that (b + d)2 is the same as b2 + d2!
They are NOT the same thing! I must evaluate the expression as it stands:
+ (4) )2 = ( 7 )2 = 49
Evaluate b2 + d2 for a = &2, b = 3, c = &4,
and d = 4.
(3)2 + (4)2 = 9 + 16 = 25
Notice that this does not
match the answer to the previous evaluation, pointing out again that exponents
do not &distribute& the way multiplication does.
Evaluate bc3 & ad for a = &2, b = 3, c = &4,
and d = 4.
(3)(&4)3 & (&2)(4) = (3)(&64) & (&8) = &192 + 8 = &184
The most common &expression&
you'll likely need to evaluate will be .
To evaluate, you take the polynomial and plug in a value for x.
Evaluate x4 + 3x3 & x2 + 6 for x = &3.
(&3)4 + 3(&3)3 & (&3)2 + 6
= 81 + 3(&27) & (9) + 6
= 81 & 81 & 9 + 6
Evaluate 3x2 & 12x + 4 for x = &2.
<font face="Times New Roman" size="3" color="#(&2)2 & 12(&2) + 4 = 3(4) + 24 + 4 = 12 + 24 + 4 = 40
Evaluate y = 4x & 3 at x = &1.
y = 4(&1) & 3 = &4 & 3 = &7
Note: This means that
the point (&1,
&7) is on the line y = 4x & 3.
Cite this article
Stapel, Elizabeth.
&Evaluation: Evaluating Expressions, Polynomials, and Functions.& Purplemath. Available from
&&&&http://www.purplemath.com/modules/evaluate.htm.
Accessed [Date] [Month] 2016 &
College Math
Standardized Test Prep
This lesson may be printed out for your personal use.
&&Copyright (C)
&&| && &&| && &&| &&Inequalities - A complete course in algebra
A L G E B R A
INEQUALITIES
< means is less than.. &This sign > means
is greater than. &In each case, the sign opens towards the larger number.
For example,& 2
2 ("5 is greater than 2").
These are the two senses of an inequality: & .
number line
On the number line, &a < b& means:& a falls to the left of b.
Problem 1.&&&Between each
pair, place the correct sign of inequality.
To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload").Do the problem yourself first!
"Or" versus "and"
The following is called a compound inequality:
x > 1 & & x & 5.
It is a compound sentence
whose conjunction is "and."& It says that x takes on values that are greater than 1 &and& less than or equal to 5.
It is within that interval that x takes its values. &The endpoint 1 is not included. &x is definitely greater than 1; we indicate that by placing a parenthesis "(". &The endpoint 5
we indicate that by placing a bracket "]".
Now consider this compound inequaltiy:
x < 1 & &x > 5.
Here, the values of x are either less than 1 &or &greater than 5.
It should be clear that x could not be a number that is less than 1 and greater than 5.& There is no such number.
So, when the conjunction is and,
x > 1 &and & x & 5.
the values of x fall inside a certain interval. &But when the conjunction is&or,
x < 1 &or &x > 5.
the values of x fall outside an interval.
Problem 2.&&&Graph these compound inequalities.
a) & &&x < &1 & or & x & 3.
b) & &&x & &1 & and & x < 3.
inequality
That is called a continued inequality.& It means
a < x &and &x < b.
A continued inequality always implies the conjunction and.& The sense is
always < &or &&.
continued inequality means:
x falls in the interval between a
and&b."a&<&x&<&b"
illustrates that.
Note: &When neither end point is included, as in this example, we call that an open interval. &When both end points are included --
-- we call that a closed interval. &(Otherwise, like the proverbial glass, the interval is half-open, or half-closed.)
Problem 3.&&&Write as a continued inequality.
a) & x > &3 &and &x < 1.
&3 < x < 1.
b) &Graph that continued inequality.
Problem 4.&&&0 < x < 6. &Write that continued inequality as a compound inequality.
x > 0 &and &x < 6.
Problem 5.&&&Write as a continued inequality:
x < 1 &or &x > 5.
Not possible! &The conjunction must be and.
Problem 6.&&&Name four values that x might have.
a) & 1 & x < 3.
For example, 1, &1.2, &2.5, &2..
b) &x < &1, &or & x > 1.
For example, &2, &&3,456,987, &1.000005, &1023.
Problem 7.&&&Write in symbols.
a) &x is a positive number. &
b) &x is a negative number. &
c) &x is a non-negative. &
x & 0. &"x is greater than or equal to 0."
theorems of inequalities
prove a statment whose predicate is "is greater than," we must have a definition of &"is greater than." &We shall adopt the following. &We shall define "a is greater than b" to mean: a & b is positive. &Algebraically:
a > b &if and only if &a & b > 0.
On the basis of this definition, we can prove various
about inequalities.
.&&& We may add the same number to both sides of an inequality, and the sense will not change.
Note: &If c is a negative number, then the theorem implies that we may
the same number from both sides. &Just as with equalites, any theorem
is also true for subtraction.
follows, from this Theorem, that we may .
implies& &
We could prove that by adding &d to both sides.
as follows:
&&&&&&&&&&&&&&&&&&&&&&
&&&&&&&&&Therefore, upon adding and subtracting c :
a & b + c & c
&&&&&&&&&&,
Which is what we wanted to prove.
Note that any theorem of inequality is true for an inequality of the opposite sense. &For, we could write: &b < a &implies & b + c < a + c.
.&&&We may multiply both sides of an inequality by the same positive number, and the sense will not change.
b, & and c > 0,
For example:
If we now multiply both sides by 3, for example, then
The sense does not change.
The proof is similar to that of . &Simply apply the .
Therefore, if c is positive, then
This means
This theorem also allows us to divide both sides by the same positive number, because division is multiplication by the .
.&&&If we multiply both sides of an inequality by the same negative number,
the sense of the inequality changes.
b, & and c < 0,
Here is an example:
If we now multiply both sides by &3, say, then
The sense changes.
(As in , this one also implies
when we divide both sides by a negative number, the sense changes, because division is multiplication by the .)
Therefore, if c is negative, then according to the :
This means
.&&&If we change the signs on both sides of an inequality, then the sense of the inequaltiy will change.
We can see that on the number line.
We could prove it simply by
&a and &b.
For example, since
&&then on changing the signs on both sides:
We may think of
as an instance of , because when we multiply or divide both sides by the same negative number, the signs on each side necessarily change.
The signs on each side have changed. &Therefore the sense also must change.
The signs changed because we divided each side by
.&&&If a, b are both positive or both negative, then on taking reciprocals, the sense of the inequality changes.
We can prove that by dividing both sides by ab (which, since a and b have the same sign, will be positive) -- but that will depend on knowing how to .
In any event, since
Problem 8.&&&Apply the theorems to complete the following with an inequality.
&&a) &If &
&&c) &If &
On dividing by a negative number, the sense changes.Equivalenly, we can think of it as changing the signs on both sides.
Problem 9.&&& Use
to prove: &If a is a positive number less than 1, then a2 is less than a.
For example, (&frac12;)2 = &frac14;, which is less than &frac12;.
implies& &
That is,& &
Problem 10.&&&Assuming that the literals all
have positive values, complete the following with an inequality.
,&&according to .
11.&&&&a > 0. &How is that possible?
a is not a number. It is a variable. It takes values which are numbers. To find the values of a such that &a > 0, apply
, and change the signs on both sides. But then we must change the sense.
implies& &
That is, since &0 = 0,
a must be a negative number. For example, if a = &5, then &a = &(&5) = +5, that is, &a is a positive number.
inequalities
A linear inequality has this standard form:
ax + b < c.
When a is positive, then solving it is identical to solving an :
As with equations, the inequality is "solved" when positive x is isolated on the left.& (The above steps follow from .)
The only difference between solving an inequality and solving an equation, is the following:
When when we multiply or divide by a negative number,the sense must change. (.)
Equivalently, when
the signs on both sides change, then the sense also must change (), as in .
Example 2.
On going to the last line, the signs on both sides changed. Therefore, the sense also changes.
The signs changed, of course, because we divided both sides by negative 2.
Alternatively,
we could immediately make 2x positive -- by changing all the signs on both sides. &But then we must also change the sense.
&2x + 5 < 11 &implies &2x & 5 > &11,
and so on.
Problem 12.&&&Solve each inequality for x. &Write a .
&&a) &5x + 3
b) &5x & 3
&&c) &&2x + 7
d) &&2x & 1
f) & 4x & 7
Problem 13.&&&In each of the following, what can you conclude about the signs of a and b?
a) &ab > 0
a and b must have the same signs. Formally,
a > 0 &and &b > 0, &or
&a < 0 &and &b < 0.
Lesson 4: &.
b) &ab < 0
a > 0 &and &b < 0, &or
&a < 0 &and &b > 0.
c) &ab = 0
Either a = 0 &or
Same as a). &a, b & 0.
Same as b). &a, b & 0.
The numerator must be 0. &a = 0.
Example 3.&&&Solve this inequality for x :
x & 3&x + 5
Solution.&&&According to , the numerator and denominator must
the same sign. &Therefore, either
1) &x & 3 > 0 &and &x + 5 > 0,
2) &x & 3 < 0 &and &x + 5 < 0.
x > 3 &and &x > &5.
Which numbers are these that are both greater than 3 and greater than &5?
Clearly, any number greater than 3 will also be greater than &5. &Therefore, 1) has the solution
x < 3 &and &x < &5.
Which numbers are these that are both less than 3 and less than &5?
Clearly, any number less than &5 will also be less than 3. &Therefore, 2) has the solution
The solution, therefore, is
x < &5 &or &x > 3.
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