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The prime factorization of 90 is 2*3*3*5. Whenever you multiply a number by itself, you double the number of times a factor appears. For example, 6=2*3 and 6^2=(2*3)^2=2*3*2*3. This means that every prime factor in a perfect square has to appear an even number of times. Since the 2 and 5 factor appear an odd number of times in 90, it can't be a perfect square.

Using formula of area of parallelogram which is base x height.

SinceX and Y are mid-point if AC and BCtherefore,AX = CX = 1/2 AC& BY = CY = 1/2 BC

AX = CY1/2 AC = 1/2 BCtherefore,AC = BC

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(80+85+90+71+60+100)/6486/681

Money in bank=₹175Money spented=25Money left=175-25=₹150

First divide no. of bushels with no. of horsethis means 1 horse consume 1.8 bushels in 36daysso 24 bushels ÷ 30 horses = 0.81.8:36::0.8:x1.8x=28.8x=16so 30 horses consumes 24 bushels in 16 days

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Ans:(d)

(80+85+90+71+60+100)/6= 486/6= 81

Total monthly income = 25000

Money she spent= (90/100)*25000= 250*90= 22500

Money she donated for socially useful causes= (3/100)*25000= 750

Therefore, money she save= 25000 - (22500 + 750)= 25000 - 23250= Rs 1750

Sec 20= 1/cos 20=1/sin 70cot 70= cos 70/sin 70cosec 20= 1/sin 20= 1/cos 70

(1+sec 20 + cot 70)=1+1/sin 70+ cos 70/sin 70take lcm(sin 70+1 + cos 70)/sin 70similarly(1-cosec 20+ tan 70)= (cos 70-1+ sin 70)/sin 70multiple both

(sin70 cos70- sin 70+sin^2 70+ cos 70-1+sin70+cos^2 70- cos 70+ sin70 cos 70)/ sin70 cos70

(2sin 70 cos 70)/ sin70 cos 70ans=2

wrong Solution

y=1 and x=1Both are one, just solve it by multiplication of matrix

y = 1x =1substitute that values u will get answer

y1x=1substitute that values you will get answer

Y=1and X=1 is right answer

x² - 2x - 8

x² +2x -4x - 8

x(x + 2) - 4 ( x + 2)

(x -4) ( x +2)

x^2 - 30x - 99= x^2 - 33x + 3x - 99= x(x - 33) + 3(x - 33)= (x+3)(x-33)

the red line is 1/3.the 2nd color line is 5/7the blue line is 7/5

and the yellow line is 3/2.

Take square both side( a + 1/a )^2 = (17/4)^2a^2 + 1/(a)^2 + 2 = 289/16subtract 4 from both sidea^2 + 1/a^2 - 2 = 225/16we can write above equation like(a-1/a)^2 = (15/4)^2soa - 1/a = 15/4.please like the solution 👍 ✔️

Total student will be 38+39+3690+17+6100+13113thanks

Total Students in class 2 is 38+36+39=113

90% of 560=90*560/100=504so dress is of 504

TwoanglesareAdjacentwhen they have a common side and a common vertex (corner point), and don't overlap.(i) yes(ii) No (RMS is inside RMQ i.e. they overlap) (iii) No (they do not have a common side) (iv) yes

19x-2+55=180; 19x=180-55+2; 19x=125+2=127

when two parallel line transverse by a line corresponding angle are equal55= 19x-255+2= 19x57/19= x3°= x

x²-4x-1

= x²-4x+4-3

= (x-2)²-(√3)²

= (x-2-√3)(x-2+√3)

So,

x = 2 +- √3

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P^2 y (1+x^2) = qx^2P^2 y= qx^2 /(1+x^2)Y = q(x/p)^2 /(1+x^2)

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square and rhombus ..

shaded region is called Alternate Angles

1/17,3/17,5/17,8/17,10/17,16/17.

answer of 1 question

1/17,3,/17,5/17,8/17,10/17,16/17.

x-y = 26

x = 3y

3y-y = 26

2y = 26

y = 13

x = 26+13 = 39

The number is 39+13 = 52

(550×(100+10))/100 = 550×110 / 100 = 55×11 = 605

no. LHS should be 15 it is not possible.

if you take above expression right you should take certain assumption.