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# If each expression under the square root is greater than or

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If each expression under the square root is greater than or  [#permalink]

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Updated on: 09 Oct 2013, 01:21
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Question Stats:

37% (01:47) correct 63% (01:39) wrong based on 1649 sessions

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If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

First, $$(2 - x)$$ has to be $$\ge 0$$ for the expression $$\sqrt{2 - x}$$ to be defined (GMAT deals only with real numbers). Therefore, $$x \le 2$$ .

Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x$$ (because $$x \le 2$$ and thus $$x - 3 \lt 0$$ ).

IMO:
$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3$$ is valid

Summing up: $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}$$.

M18-16

Originally posted by study on 21 Nov 2009, 00:40.
Last edited by Bunuel on 09 Oct 2013, 01:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If each expression under the square root is greater than or  [#permalink]

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09 Oct 2013, 01:22
21
38
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x\geq{0}$$ --> $$x\leq{2}$$. Next: as $$x\leq{2}$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Answer: A.
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Re: Inequalities - M08  [#permalink]

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21 Nov 2009, 01:39
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study

Quote:
Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

yes, you can and in fact should change it to (3-x). Here's explanation:

http://en.wikipedia.org/wiki/Square_root

$$sqrt(x^2) = |x|$$

|x| = x, for all x>=0
|x| = -x, for all x<0

In our case we have $$\sqrt{(x-3)^2}=|x-3|=-(x-3)=3-x$$, because (2-x)>=0; x<=2; (x-3)<0

And so the correct answer is

$$3-x+\sqrt{2-x}+x-3=\sqrt{2-x}$$
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Re: Inequalities - M08  [#permalink]

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08 Oct 2013, 15:16
1
study wrote:
Can someone please explain the logic here. This problem is from M08.

If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$ ?

* $$\sqrt{2 - x}$$
* $$2x - 6 + \sqrt{2 - x}$$
* $$\sqrt{2 - x} + {x - 3}$$
* $$2x - 6 + \sqrt{x - 2}$$
* $$x + \sqrt{x - 2}$$

First, $$(2 - x)$$ has to be $$\ge 0$$ for the expression $$\sqrt{2 - x}$$ to be defined (GMAT deals only with real numbers). Therefore, $$x \le 2$$ .

Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x$$ (because $$x \le 2$$ and thus $$x - 3 \lt 0$$ ).

IMO:
$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3$$ is valid

Summing up: $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}$$.

What's the OA? I chose (B) I'm afraid I might be wrong though
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Re: Inequalities - M08  [#permalink]

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08 Oct 2013, 20:48
21
9
jlgdr wrote:
study wrote:
Can someone please explain the logic here. This problem is from M08.

If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$ ?

* $$\sqrt{2 - x}$$
* $$2x - 6 + \sqrt{2 - x}$$
* $$\sqrt{2 - x} + {x - 3}$$
* $$2x - 6 + \sqrt{x - 2}$$
* $$x + \sqrt{x - 2}$$

First, $$(2 - x)$$ has to be $$\ge 0$$ for the expression $$\sqrt{2 - x}$$ to be defined (GMAT deals only with real numbers). Therefore, $$x \le 2$$ .

Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x$$ (because $$x \le 2$$ and thus $$x - 3 \lt 0$$ ).

IMO:
$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3$$ is valid

Summing up: $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}$$.

What's the OA? I chose (B) I'm afraid I might be wrong though

The correct answer will be (A)
The expression becomes

$$|x-3| + \sqrt{2 - x} + (x - 3)$$

Note that $$\sqrt{x^2} = |x|$$ (and not x)

Since we know that x < 2, (x - 3) must be negative. Hence $$|x-3| = -(x - 3)$$

Expression becomes: $$-(x - 3) + \sqrt{2 - x} + (x - 3) = \sqrt{2 - x}$$
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Re: If each expression under the square root is greater than or  [#permalink]

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09 Oct 2013, 17:29
Bunuel wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Answer: A.

I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?
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Re: If each expression under the square root is greater than or  [#permalink]

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09 Oct 2013, 19:27
2
3
AccipiterQ wrote:
Bunuel wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Answer: A.

I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?

Absolutely not! Ordinarily, if you change the order, it changes everything. But this situation is not ordinary. It is a mod situation i.e. there is modulus around x - n.

Recall how mods are defined:
|x| = x if x is positive
= -x if x is negative

So if you want to get rid of the mod sign, you need to know whether x is positive or negative.

Here we know that x <= 2. For all such values of x, (x-3) is negative.
So |x-3| = -(x - 3) = 3 - x
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Re: If each expression under the square root is greater than or  [#permalink]

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10 Oct 2013, 05:57
VeritasPrepKarishma wrote:
AccipiterQ wrote:
Bunuel wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Answer: A.

I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?

Absolutely not! Ordinarily, if you change the order, it changes everything. But this situation is not ordinary. It is a mod situation i.e. there is modulus around x - n.

Recall how mods are defined:
|x| = x if x is positive
= -x if x is negative

So if you want to get rid of the mod sign, you need to know whether x is positive or negative.

Here we know that x <= 2. For all such values of x, (x-3) is negative.
So |x-3| = -(x - 3) = 3 - x

OOOOH I see now, thank you,
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Re: If each expression under the square root is greater than or  [#permalink]

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05 Nov 2013, 08:03
WOW! That is a really well concealed trap.
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Re: If each expression under the square root is greater than or  [#permalink]

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06 Jul 2014, 20:55
Bunuel wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x\geq{0}$$ --> $$x\leq{2}$$. Next: as $$x\leq{2}$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Answer: A.

Hi,
As per GMAT Club Math book v3 page 14 - Roots... It is mentioned " • When the GMAT provides the square root sign for an even root, such as or , then the only accepted
answer is the positive root."

So how can one take |x-3| as the square root? The square root can only be +ve (x-3)
Please explain.
Thanks
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Posts: 64125
Re: If each expression under the square root is greater than or  [#permalink]

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07 Jul 2014, 00:27
nehamodak wrote:
Bunuel wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x\geq{0}$$ --> $$x\leq{2}$$. Next: as $$x\leq{2}$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Answer: A.

Hi,
As per GMAT Club Math book v3 page 14 - Roots... It is mentioned " • When the GMAT provides the square root sign for an even root, such as or , then the only accepted
answer is the positive root."

So how can one take |x-3| as the square root? The square root can only be +ve (x-3)
Please explain.
Thanks

An absolute value is always more than or equal to zero. So, |x - 3| >=0.
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Re: If each expression under the square root is greater than or  [#permalink]

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07 Jul 2014, 04:12
nehamodak wrote:
Bunuel wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x\geq{0}$$ --> $$x\leq{2}$$. Next: as $$x\leq{2}$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Answer: A.

Hi,
As per GMAT Club Math book v3 page 14 - Roots... It is mentioned " • When the GMAT provides the square root sign for an even root, such as or , then the only accepted
answer is the positive root."

So how can one take |x-3| as the square root? The square root can only be +ve (x-3)
Please explain.
Thanks

Let me also add here that when we take absolute value of a number/expression, we are implying that it doesn't matter whether the number is positive or negative - we will take it to be positive only.
Say, if x-3 is 7, then |x-3| is 7. When x-3 is -7, |x-3| is 7 only.
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Re: If each expression under the square root is greater than or  [#permalink]

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22 Jul 2015, 00:22
1
If we just plug in the number 1 we can get the answer in 30 seconds max.
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Re: If each expression under the square root is greater than or  [#permalink]

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23 Sep 2015, 10:41
1
i assumed x=1 and compared the answer with the options and A is only matching.
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Re: If each expression under the square root is greater than or  [#permalink]

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09 Jan 2016, 17:16
How should we intuitively know that x is = or < 0?

Would a correct approach include setting x to 0 and plugging in to derive an answer, then setting x to a negative number say -2 and plugging in to derive an answer, and then picking the corresponding answer from the answer choices? Or am I way off here?
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Re: If each expression under the square root is greater than or  [#permalink]

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10 Jan 2016, 05:27
Anonamy wrote:
How should we intuitively know that x is = or < 0?

Would a correct approach include setting x to 0 and plugging in to derive an answer, then setting x to a negative number say -2 and plugging in to derive an answer, and then picking the corresponding answer from the answer choices? Or am I way off here?

Please re-read the solutions above. We derived that x≤2 not x≤0.
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Re: If each expression under the square root is greater than or  [#permalink]

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10 Jan 2016, 10:23
Ah, thank you, I see how the condition on x was derived now.
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If each expression under the square root is greater than or  [#permalink]

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08 Mar 2018, 16:16
study wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

Main Idea: We are likely looking for cancellation of terms

Details: We see the first term is sqrt(x-3)^2 and looks like the right term for cancellation with (x-3)

sqrt(x-3)^2 = |x-3| which can be x-3 or -(x-3) = -x+3

We can rule out x-3 because x< 2 and so it is negative

We now have the expressions as : -X+3 + sqrt(2-x) +x-3 = sqrt(2-x)

Hence A
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Re: If each expression under the square root is greater than or  [#permalink]

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30 Jun 2018, 08:12
Hi Bunuel, chetan2u, VeritasPrepKarishma
I used the following approach, not sure if its the right reasoning though, can you please check it out
x^2-6x+9=0
D=(6^2)36-4*9=0
x=(6+-0)=3,
so x is 3. substitute x with 0 in the expression and we get A as correct answer only.
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Re: If each expression under the square root is greater than or  [#permalink]

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30 Jun 2018, 09:09
nigina93 wrote:
Hi Bunuel, chetan2u, VeritasPrepKarishma
I used the following approach, not sure if its the right reasoning though, can you please check it out
x^2-6x+9=0
D=(6^2)36-4*9=0
x=(6+-0)=3,
so x is 3. substitute x with 0 in the expression and we get A as correct answer only.

No, this is not right at all.

We don't have an equation here. We have an expression. From where did you get that x^2 - 6x + 9 = 0? And if x = 3, why are you substituting x = 0, then?
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Re: If each expression under the square root is greater than or   [#permalink] 30 Jun 2018, 09:09

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