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nB/1-tanAtanBWhen A=π/6 and B=π/3put the valuetanπ/6 tanπ/3 /1- tanπ/6 tanπ/31/√3 ×√3   1-1/√3×√31/1-1⇒1/0=∞=which cannot be DEFINED

is : 51/303/5 * 2 5/6 = 3/5 * 17/6 = 51/30

/5 = 12/25 which is the SIMPLEST  so ANS is a 12/25

and  y = x³these curve meet  when  x³ = √x       =>  x = 0  or  1y = √x  is higher than  x³  between x= 0 and x = 1.areas under the curves / enclosed by the curves :==============================at any x, the radius of the circular cross section of the paraboloid is y..Volume of an infinitesimal  volume of a disc of radius y  and  thick ness   dx       = π y²  dx   the equation of the curve is     y = √x      or  x =  y²Volume of the paraboloid :=====================R = 50.0 ft  A = 2 π / 3 radians = 120 deg. = 2π/3 radians Height raised by, when the wheel turns by 90 deg.  =  1 radius. = 50 ft.           as the person is at the level of the center of the wheel.height raised by, when the wheel turns by 30 deg more =  R Sin (2π/3 - π/2) = R/2 = 25 ft.The height above the ground = 25 + 50 = 75 feet..======================ANGULAR speed = ω = angle turned / time DURATION = [ 2 π /3 ] / 30 sec              = π / 45  radians / sec              = 0.0698 radians / sec

3=3abso, a+b+c= AB

he triangle is ISOSCELES, the other two sides have EQUAL length. Call the length of each side .DRAW a perpendicular from vertex to the base and recall that this is also a bisector of base. Therefore you get a right triangle with a leg and one angle . The other side can be obtained by applying cosine ratio :Add up the sides to get the perimeter

system has TWO types -                                                                         (i). NATURAL number                                                                                             (II). WHOLE number .

is 21-3/6( in WICH 21is on RIGHT SIDE)

to Solvewe have to CONVERT 2/10 to  20/100then the FRACTION BECAME 59/100 + 20 / 100Therefore  79/100

5 odd NUMBERS can never be even. its POSIBLE only in the case of FACTORIAL....3!+3!+3!+3!+3! = 30hope its HELP u......

eBased on Angles : 1)Acute angled triangle - All the  angles are acute angle2)Right angled triangle - 1 angle is right angle3)Obtuse angled triangle - 1 angle is  obtuse angleBased on sides :1)Equilateral triangle - All sides and angles are equal2)Isosceles  triangle - Only 2 sides are equal and angles OPPOSITE to the  equal sides are equal3)Scalene triangle - No Sides are equalQuadrilateral 1) Trapezium - Diagonals intersect each other.                       Line joining the midpoints of the non parallel sides is parallel                         to EITHER of th two parallel sides .2) Parallelogram [ Square , Rectangle and Rhombus come under parallelogram] - Opposite sides and angles are equal .                          Diagonals of a parallelogram bisect each other .3) KITE - Has 2 pairs of adjacent  sides  equal .

raw a normal number .DRAW a 1 CM line above 3 and name the point B and join it to 0.take a compass and measure this line segment and draw an ARC from 0 cutting the number line.this point is root 10.REASON:HYPOTENUSE square =sum of the squares of other two sides=square of 3+ square of 1= 9+1=10hypotenuse= root 10

daily life DATA handling is used in different way.data handling is used to TALLY marks,to record the temperature,to CACULATE literacy rate of state,to record the net attendance rate of SCHOOL students e.t.c.

bers don't add up to give you an even number.It is POSSIBLE only in case of factorial.3! + 15 + 9 = 30

a horizontal line BD = 13 cm ii) At B draw an angle DBX = 75 deg [Angle may be constructed with protractor or with compass] iii) In BX cut off an arc of radius = 7 cm; let this point be C iv) Join CD v) Using compass draw a perpendicular bisector for CD, which meets BD at A.vi) Join AC; ABC is the REQUIRED triangle. Reason: SINCE A is a point on the perpendicular bisector of CD, by locus theorem it is equidistant from the end points C & D of the line segment CD. Hence AC = AD. Thus BD = BA + AD = BA + AC = 13 cm

e any ray OA. (ii) With O as centre and any convenient RADIUS, DRAW an ARC cutting OA at B. (iii) With B as centre and the same radius, draw an cutting the FIRST arc at C. (iv) With C as centre and the same radius, cut off an arc cutting again the first arc at D. (v) With C and D as centre and radius of more than half of CD, draw two arcs cutting each other at E, join OE.

ollow cylinder is FORMED and the length will be the circumference and the BREADTH will be height

be DONE by USING FACTORIAL 3!+9+15

of PUPIL after DECREASING = 900 - 1/3*900 = 600

I THINK it is 90....

its the HCF of these 3 NUMBERS and it is 7

perpendiculars from M MEET AB &AC at X and y respectivelygivenAP bisects

4029×2 + 9×2246st = 8058+18         = 8076

a few math PROBLEMS:Q: A SOCCER field has an area of 7/8 SQUARE MILES. The length of the field id 2/5 miles. What is the with of the soccer field?A: 35/16Q: Apples aer 5 for 1.50 and bananas are 6 for 1.20. What is the total cost of buying 2 bananas and 2 apples?A: 1.00

/301-296+20=315/5+20=315/25=12.6

total no. of questions= 25Marks scored= 80%THEREFORE, Absolute marks scored= 80% of 25= 80/100 * 25= 80/4=20Therefore, 20 questions were ANSWERED correctly.

1-sinA)(SECA + TANA)= (1/cosA)(1-sinA)(1/cosA + sinA/cosA)= (1-sinA/cosA)(1+sinA/cosA)= 1-sin²A/cos²A= cos²A/cos²A= 1Hope that HELPS !!

+ sec A)² + (Cos A + COSEC A)² = Sin² A + sec² A + 2 SIN A Sec A + Cos² A + Cosec² A + 2 Cos A Cosec A= Sin²A + COs²A + Sec²A + Cosec²A + 2 Sin A SECA + 2 Cos A Cosec A= 1  +  [ 1/Cos²A + 1/ Sin²A ] +  [ 2 Sin A / Cos A + 2 Cos A / SIn A ]= 1 + (Sin²A + COs²A)/ [Cos²A Sin²A ]  + 2 [ SIn² A + Cos²A ] / [ SinA CosA= 1 + 1/Cos²A 1/Sin²A  + 2 1/SinA  1/CosA= 1 + Sec²A Cosec²A + 2 COsecA Sec A= (1 + SecA CosecA )²====================alternately,(sin A + sec A)² + (Cos A + Cosec A)² =  (Sin A + 1/Cos A)² + (COs A + 1/ SinA)²=   (Sin A Cos A  + 1)² / Cos² A    +  (SinA COsA + 1)² / Sin² A= [ SIn A Cos A + 1]² [ 1/Cos² A + 1/Sin² A ] = [  SIn A Cos A + 1]² [ Cos² A + Sin² A ] / [Sin²A Cos² A ]= [ Sin A Cos A + 1 ]²  /  [Sin²A Cos² A ]= [ (Sin A Cos A + 1) / (Sin A Cos A )]²=  (1 + 1/Sin A  1/Cos A)² = (1 + Sec A Cosec A)²

on of CUBOIDAL tin = 30cm×40cm×50cmlet length = 40cmbreath = 30cmheight = 50cmTotal SURFACE area = 2(lb+bh+lh)                           = 2(40×30 + 30×50 + 40×50)                           = 2(1200 + 1500 + 2000)                           = 2(4700)                           = 9400 cm²TSA of one tin box = 9400 cm²TSA of 20 tin BOXES = 9400×20 = 18800 cm² = 1.88m²Cost of tin = Rs. 20/m²Total cost = 1.88×20 =Rs. 37.60

hat , ∠2 = ∠3 ( AP is a bisector ) const .     draw M ⊥  on AP and join to AB on P and similarly in other SIDE M ⊥ AP and join that to AC at Q . we get ,   ∠1 = ∠4 = To prove -  PM = QMproof -  in Δ APM and Δ AMQ        ∠2 = ∠3 ( GIVEN  )      AM = AM ( COMMON )     ∠1 = ∠4 ( by const. ) by ASA Δ APM congruent to Δ AMQ .⇒ PM = QM ( C.P.C.T)

I don't knowStep-by-step EXPLANATION:OK NAI am SORRY

Step-by-step explanation:52000*6/`100