Explore topic-wise InterviewSolutions in .

-0.32 is the answer of the given question

-0.32 is the correct answer

area=1/2(12+8)(4)+(12)(6)+1/2(12+8)(4) =40+72+40=152m^2

wrong answer

hit like if you find it useful

Triangle ABD is a right triangle,SoBD^2=AB^2+AD^2=BD^2=(40m)^2+(9m)^2BD^2=1600m^2+81m^2=1681m^2BD=root1681or,BD=4cmarea of triangle ABD=1/2×9m×40m=180m^2In triangle TBD ,area = 1/2×30×13=195m^2 area of quadrilateral = 180+195+195=570 m^2

As we knowX+y+z= 360°Z= 28°And x=ySoY+y+28= 3602y= 360-282y= 332Y= 332/2= 166°Now x= 166°Please like the solution 👍 ✔️👍

126 is lcm of 42,63 by division method...

126 is correct answer.

Cube root of 7532 can be expressed in symbol as∛7532

∛7532= 19.6021377649

X+20 = 2x[angle at centre = 2* angle at circumference]=> X = 20.

Please hit the like button if this helped you

Let Rohan's age be x Therefore Rohan's mom's age = x+26. Now Rohan age after 3 yrs =x+3 and his mothers age after three years x+29...... Now we know (x+3)(x+29) = 360 . x²+3x+29x+87 = 360 . x²+32x+87=360 . x²+32x=360-87 = 273 x(x+32)=273 After solving we get x=7 So, age of Rohan = 7 yrs The solution for how to get x is Take one factor of the equation x²+32x-273 = 0 First factorise .... Factors are x-7 and x+32 ...... So take any one factor and solve it and you will get two solutions of the quadratic equation...... X-7.....x=7 And x+32.....x=-32 Now put your logic age can't be in minus so x=7 is correct solution for this question...

hit like if you find it useful

Solution: Area of shaded part is 693cm^2

Explanation:Area of shaded part will be as we can see there are 4 half circle that makes 2 circles

Side of square = dnow d will be diameter of CircleNow (d/2) + d + (d/2) = 42 2d = 42 d = 21 cmSo radius will be (21/2) cm

Shaded area = 2* π r^2 = 2 * (22/7) * (21/2) * (21/2) = 2 * (22/7) * (110.25) = 693cm^2 will be the area of shaded part

HIT THE LIKE BUTTON ✔️👍

Tell me if you have any doubt

PN can be expressed asPN = MN - NP

Number of edges:12

Number of faces:6

The figure is a cube.

the answer could be 490

ans is 99 it's very simple

the correct answer is 99

Ans :- The longdivision methodis used when you are dividing a large number (usually three digits or more) by a two-digit (or more) number. It is set out in a similar way to shortdivision(the 'bus stop'method).

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)Let’s consider the ratio these two AP’s mth terms as am : a’m →(2)Recall the nth term of AP formula, an = a + (n – 1)dHence equation (2) becomes,am : a’m = a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1 : S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].

We get22.360up to three decimal places.

Explanation:

As the given number is not a perfect square, the square root of this number will be a decimal number.We must, therefore, state the accuracy to which the square root is to be calculated.

Suppose we need to calculate the√500to an accuracy of three decimal places.

Step 1. Make pairs of digits starting from Units place.we get5¯¯¯¯00.

Step 2. Consider the unpaired digit and calculate the nearest digit which is just lower than the square root of this digit. We consider5.The nearest digit is2as32=6>5.

We get first digit of the answer. Workout the remainder which=1. Bring down the next pair which is¯¯¯¯00. Makes the dividend=100. Also add the answer digit to the divisor. We have2+2=4

Step 3. Place highest digit in units place of divisor to make the divisor which will divide dividend100. We see that it can be divided with2×42.∴3×43>100.Complete the division. Nearest digit is2. Complete the division and obtain remainder.=100−84=16. Also add the answer digit so obtained to divisor. We get42+2=44for the next step.

Step 4. Next we have decimal in the original number. Place it after the quotient22and bring a pair of zeros down. The dividend becomes1600

Continue steps till three places of decimal.

We see that in the Step 5. Number remaining is304.Adding two zeros makes it30400We are not able to divide the dividend because even with digit1placed at units place of divisor the product of quotient digit and divisor is>the dividend. We place a0in the quotient place and add two zeros to the dividend.

and where is that adjacent figure ??

Plz suggest what to do

recall that: sinA + sinB = 2sin[(A + B)/2] cos[(A - B)/2]

0 ≤ x < 2π then 0 ≤ 2x < 4π 0 ≤ 4x < 8π

sin(2x) + sin(4x) + sin(6x) = 0 let y = 2x we have: siny + sin2y + sin3y = 0 sin3y + siny + sin2y = 0 2sin[(3y + y)/2] cos[(3y - y)/2] + sin2y = 0 2sin2y cosy + sin2y = 0 sin2y (2cosy + 1) = 0 sin2y = 0, cosy = -1/2 sin4x = 0, cos2x = -1/2 4x = 0, π, 2π, 3π, 4π, 5π, 6π, 7π OR 2x = 2π/3, 4π/3, 8π/3, 10π/3 x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4 OR x = π/3, 2π/3, 4π/3, 5π/3 x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, π/3, 2π/3, 4π/3, 5π/3 x = 0, π/4, π/3, π/2, 2π/3, 3π/4, π, 5π/4, 4π/3, 3π/2, 7π/4, 5π/3 ⇦ ANSWER

1

2

4 is where

ah s h. y8 dn ayd saakey

statement is where

V = (4/3)πr³So the volume of a hemisphere will be half of that, which is (2/3)πr³

The volume of a cone of base radius r and height h is:V = (1/3)πr²h

We are told that r = 4.2 cm

The total height of the toy is made up of (the height of the cone) + (height of hemisphere)So the height of the cone = 10.2 - 4.2 = 6 cm

So the volume of the toy = (volume of hemisphere + volume of cone)= (2/3)π * 4.2³ + (1/3)π * 4.2² * 6= 155.2 cm³ + 110.8 cm³= 266 cm³

Like if you find it useful

S6 =42

a + 9d 1------------ = -------a + 29d 3

cross multiply we get

3a + 27d = a +29 d

2a - 2d = 0 ------------------ (1)its given thatsum of first six terms of an AP is 42

therefore

S6 = n/2 ( 2a + (n-1) d)

42 = 6/2 ( 2a + (6-1) d)

42 = 3 (2a + 5d )

14 = 2a +5d

2a +5d = 14 ------------------- (2)

solve eq 1 and 2

2a - 2d = 02a +5d = 14

we getd= 2a = 2---------------13th term of AP= a + (n-1)d

2+ (13-1) 2

= 2+ 24

=26

Like my answer if you find it useful!

Let a1 = the first term....we have....the sum of the 1st 6 terms = 42....therefore...

42 = (6/2)[2a1+ 5d]

42 = 3[2a1+ 5d]

14 = 2a1+ 5d → a1= [14 - 5d]/2

And a30= 3(a10) .... therefore.....

3(a10) = a10+ 20d

2(a10) = 20d

a10= 10d

So we have

a10= (a1+ 9d) ... and by substitution.....

10d = ([14- 5d]/2 + 9d)

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a1= [14 - 5(2)] / 2 = [14 - 10] / 2 = 4/2 = 2

And

a13= [2 + 2(12)] = 26

S6 =42

a + 9d 1------------ = -------a + 29d 3

cross multiply we get

3a + 27d = a +29 d

2a - 2d = 0 ------------------ (1)its given thatsum of first six terms of ap is 42

therefore

S6 = n/2 ( 2a + (n-1) d)

42 = 6/2 ( 2a + (6-1) d)

42 = 3 (2a + 5d )

14 = 2a +5d

2a +5d = 14 ------------------- (2)

solve eq 1 and 2

2a - 2d = 02a +5d = 14

we getd= 2a = 2---------------13th term of AP= a + (n-1)d

2+ (13-1) 2

= 2+ 24

=26

25-5×2+3-8÷2 is the right answer

25-5×2+3-8÷2 is the correct answer

a1=1 a2=1 a3=1 a4 = a3+a2 = 1 + 1 = 2a5 = a3+a4 = 2 + 1 = 3a6 = a4+a5 = 2 + 3 = 5

I don't know this very hard

π\6 is the right answer

Pai /6 is correct answer

In ∆ ABC and ∆BDCAB = BDAC = CDBC= BCby SSS rule ∆ ABC and ∆BDC are congurent

We know that sin²x+cos²x = 1

So sin²29 + cos²29 = 1

Like my answer if you find it useful!