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In series 120, 320, ? , 2070, 5196, 13007.5

Difference between successive terms is given by

320 = 120*2.5 + 20

So, Third term= 320*2.5 + 20= 160*5 + 20= 820

Perimeter is sum of all sideshence2/3x+x+1/X+6/x^2

Let the no be xSo-240/x=16-240=16xX=-240/16X=-15 ans.

let that no is x -240/x=1616x= -240x= -240/16= -15that no is -15

suppose _no is x,

-240/x = 16

-240 = 16 x

x = -240/15x = -15

the correct answer of this question is -15

3x-15+x+10+2x-5, 5x=15+5-10; 5x=10; x=10/5=2

3x-15+2x-5+x+10=180; 3x+2x+x+10-20=180; 6x=180-10=170; x=170/6=28.333

AUB= all NUMBERS, A(~)B= different A

isme bahut mehnat ki hai please like to Kar do

AUB=all numbers,A(~)B=different A

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Volume of hemisphere=2/3*pi*r^3Volume of right cone = 1/3*pi*h*r^2

2Vh = Vc(Vh = volume hemisphere, Vc = volume cone)

4/3*pi*r^3=1/3*pi*h*r^2multiply by 3

4pi*r^3=pi*h*r^2

If they have the same base, their radii are the same. dividing both sides by pi*r^2 yields

4*r=h

2r=diameter, so r=d/2

4*d/2=h

2d=hhence ratio is 1:2

given h=10cmr=5cmV=(pi)r^2 hV=3.14×5×5×10V=785cm cube

Let the required radius be rVolume of hemisphere+ Volume of cone = Volume of the solid⇒2/3​πr3+ 1/3​πr2h = 32/3​π⇒2r3+ 4r2= 32⇒r3+ 2r2= 16⇒r3+ 2r2- 16 = 0By replacing r with 2,we get the resultant as 0Therefore, the required radius is 2cm

y-9=21y=21+9y=30.

the answer is 4r and it's power 3

4r³ is the correct answer of the given question

use (a+b)^2= a^2+b^2 +2abso given eq reduce to(p+3)^2 - 9r^2so (p+3-3r)(p+3+3r)

The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .

now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .

we also know, General equation of circle is x² + y² + 2gx + 2fy + C = 0

when point (0,0) (0)² + (0)² + 2g(0) + 2f(0) + C = 00 + 0 + 0 + 0 + C = 0 C = 0 -------(1)

when point (a,0) (a)² + (0)² + 2g(a) + 2f(0) + C = 0a² + 2ag + C = 0from equation (1) a² + 2ag = 0 a(a + 2g) = 0g = -a/2

when point ( 0, b)(0)² + (b)² + 2g(0) + 2f(b) + C = 0b² + 2fb + C = 0f = -b/2

now, equation of circle is x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }x² + y² -ax -by = 0

wrong

volume = πr²hwhen h= rit will be πr³

Perpendicular distance from y-axis = x coordinate of the pointGiven x coordinate of point is 4.5,Therefore, perpendicular distance of point A(4.5,13)from y-axis is 4.5

9 : 7 : 3Total units = 19 units19 u = 266 inches1 u = 266 / 19 = 14 inchesShortest side = 3 u3 u = 14 x 3 =42 inches

The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .

now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .

we also know, General equation of circle is x² + y² + 2gx + 2fy + C = 0

when point (0,0) (0)² + (0)² + 2g(0) + 2f(0) + C = 00 + 0 + 0 + 0 + C = 0 C = 0 -------(1)

when point (a,0) (a)² + (0)² + 2g(a) + 2f(0) + C = 0a² + 2ag + C = 0from equation (1) a² + 2ag = 0 a(a + 2g) = 0g = -a/2

when point ( 0, b)(0)² + (b)² + 2g(0) + 2f(b) + C = 0b² + 2fb + C = 0f = -b/2

now, equation of circle is x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }x² + y² -ax -by = 0

Diagram for the question

We have a straight line AB.Take a point P not on AB, but some distance away. We draw PC the perpendicular on to AB meeting AB at C. We draw another line from P to D, where D lies on AB. We show that PD > PC, for any point D other than C.

In the triangle PCD, use the Pythagorean law for sides:PC² + CD² = DP²

Since CD² is positive and adds to PC², PC²+CD² > PC²Hence DP² > PC²

So DP > PC

SO PC, the perpendicular distance from an external point to a line segment, is the shortest

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The sum of n terms in an AP is given bys= n/2(2a+(n-1)d)wheres = sum of progressionn = number of termsa = value of first termd = difference between 2 terms

now, the first term in your question is 3 (2 x 1 + 1)so, a = 3second term is 5

third is 7, like that

so d = 2

Therefore, the sum of first n terms of the AP will be

n/2 [2 x 3 + (n-1) 2]simplifying, we get the sum as

n [3+n-1]

n (n+2)Hencen=33(3+2)=3*5=15

area of circles is 22/7×5×5=78.5714 is the right answer .....

78.5 is the right answer.....

the correct answer is 78.5

lcm = a²bq

hcf = abq

two of them are rhombus (false) as one is kite shaped

thank you

no your answer is wrong

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9x/7-6x=159x=105-90x90x-9x=10581x=105x=105/81x=35/27

2x-7x/1-5x=3+7x/4-5x-5x/1-5x=3+7x/4-5x-20x+25x*2=3-35x*235x*2+25x*2-20x-3=060x*2-20x-3=0

20+_√400-720X=--------------------------- 2×60

X=20+_√-320 --------------- 120 I hope it will help you👉👉👤

Tq u

we know that sum of parallel sides is equal to twice of the line joining the mid points of non parallel sides.

you can use the mid point theorem and its conversefor solving the question.in triangle ACP,X is mid point of PC [converse of mid point theorem]also BX =1/2 AP [mid point theorem].(1)​similarly by using both theorems in triangle CRP,we can proveXQ=1/2 CR .(2)(1)+(2)= BX +XQ= 1/2(AP+CR)​this implies AP+CR=2 BQ