Answer:
1+1

=5-2+2-3

=3-1

=-1+3

=5-4+3-3+1

=2+2+2-4

=2

x+4x=y+4yx+4x=y+4y

⇔x2+4x=y2+4y⇔x2+4x=y2+4y

⇔y(x2+4)=x(y2+4)⇔y(x2+4)=x(y2+4)

⇒x2y+4y=xy2+4x⇒x2y+4y=xy2+4x

⇒x2y−xy2=4x−4y⇒x2y−xy2=4x−4y

⇒xy(x−y)=4(x−y)⇒xy(x−y)=4(x−y)

⇒xy(x−y)−4(x−y)=0⇒xy(x−y)−4(x−y)=0

⇒(xy−4)(x−y)=0⇒(xy−4)(x−y)=0

But x ≠ y 

⇒xy−4=0⇒xy−4=0

⇒xy=4

From the question, we deduce that 18 is the third integer. Let 18 - 2a ; 18 - a ; 18 + a ; 18 + 2a be the remaining integers. We have :

(18−2a)2+(18−a)2+182+(18+a)2+(18+2a)25=374(18−2a)2+(18−a)2+182+(18+a)2+(18+2a)25=374

⇔182−72a+4a2+182−36a+a2+182+182+36a+a2+182+72a+4a25=374⇔182−72a+4a2+182−36a+a2+182+182+36a+a2+182+72a+4a25=374

⇔182.5+10a25=374⇔182+2a2=374⇔2a2=50⇔a2=25⇔a=±5⇔182.5+10a25=374⇔182+2a2=374⇔2a2=50⇔a2=25⇔a=±5

The answer is : 18+5×2=28

Let x, y, z be the number of marbels each person has.

After Xavier shared the marbles, Yvonne had 2y, Zeena had 2z and Xavier had x - y - z.

After Yvonne shared the marbles, Xavier had 2(x - y - z), Zeena had 4z and Yvonne had : 2y - (x - y - z) - 2z = 3y - x - z

After Zeena shared the marbles, Xavier had 4(x - y - z), Yvonne had 2(3y - x - z) and Zeena had : 4z - 2(x - y - z) - (3y - x - z) = 7z - x - y

We have :

⎧⎪⎨⎪⎩4(x−y−z)=482(3y−x−z)=487z−x−y=48⇒⎧⎪⎨⎪⎩x−y−z=12−x+3y−z=24−x−y+7z=48{4(x−y−z)=482(3y−x−z)=487z−x−y=48⇒{x−y−z=12−x+3y−z=24−x−y+7z=48⇒⎧⎪⎨⎪⎩x=78y=42z=24⇒{x=78y=42z=24

The answer is : x−4(x−y−z)=78−4.(78−42−24)=78−4.12=30

Draw CH⊥ABCH⊥AB. 

ΔABCΔABC right at C has AB=√AC2+BC2=√122+92=15AB=AC2+BC2=122+92=15 

AC2=AH.AB⇒AH=12215=9.6⇒DH=9.6−5=4.6AC2=AH.AB⇒AH=12215=9.6⇒DH=9.6−5=4.6

ΔACHΔACH right at H has CH=√AC2−AH2=√122−(9.6)2=7.2CH=AC2−AH2=122−(9.6)2=7.2

ΔCDHΔCDH right at H has

CD=√CH2+DH2=√(7.2)2+(4.6)2=√73CD=CH2+DH2=(7.2)2+(4.6)2=73 (units)

Let a and 1.5 a be the width and the length of the swimming pool

a x 1.5a = 216⇔a2=144⇔a=12⇔1.5a=18⇔a2=144⇔a=12⇔1.5a=18

The outside length of the deck is : 18 + 5 + 5 = 28 (ft)

The outside width of the deck is : 12 + 5 + 5 = 22 (ft)

The answer is : 2(28 + 22) = 100 (ft)

Let O be the center of the circle, then OA = OB = OC = AB

⇒ΔOAB⇒ΔOAB is an equilateral triangle ⇒ˆAOB=600⇒AOB^=600

ˆO1O1^ is the exterior angle of ΔAOCΔAOC isosceles at O, so ˆO1=2ˆC1O1^=2C1^

Similarly, we have ˆO2=2ˆC2O2^=2C2^

ˆACB=ˆC1+ˆC2=12(ˆO1+ˆO2)=12.600=300

There are 9 choices to choose the first digit 

There are : 10 - 1 = 9 choices to choose the second digit 

There are : 10 - 2 = 8 choices to choose the third digit 

There are : 10 - 3 = 7 choices to choose the last digit 

The answer is : 9 x 9 x 8 x 7 = 4536 (numbers)

Let a be the number of jelly beans in the jar.

After Kathy ate, the jar had a−18a=78aa−18a=78a.

Sue ate : 78a×15=740a78a×15=740a

After Sue ate, the jar had : 78a−740a=710a78a−740a=710a

Kathy and Sue ate : a−710a=310aa−710a=310a

Pat ate : 310a×2=35a310a×2=35a

Drew ate : 710a−35a=110a710a−35a=110a

The answer is : 110a:35a=16

Let y = ax + b be the eqaution of the line. We have :

{2a+b=04a+b=−3⇒{2a=−3b=−2a{2a+b=04a+b=−3⇒{2a=−3b=−2a⇒⎧⎨⎩a=−32b=3⇒{a=−32b=3

So, the equation of the line is y=−32x+3y=−32x+3

The answer is : (0 ; 3)

Denote AB = a, then SABC=a2√34;AD=DB=a√2SABC=a234;AD=DB=a2

⇒SABD=a24⇒SACBD=a2(√3−1)4⇒SABD=a24⇒SACBD=a2(3−1)4

The answer is : √3−1≈0.73

The center of the circle is the midpoint of the diameter. Its coordinates are : (−3+112;−2−102)=(4;−6)

The sum of the scores of 14 quizzes is : 81 x 14 = 1134

The sum of the scores of first 10 quizzes is : 75 x 10 = 750

The sum of the scores of last 4 quizzes is : 1134 - 750 = 384

The answer is : 384 : 4 = 96