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P(x) = x^2 + x - 20

x^2 + 5x - 4x - 20 = 0x(x + 5) - 4(x + 5) = 0(x - 4)(x + 5) = 0x = 4, - 5

Therefore,Zeroes of given polynomial are 4 and - 5

The coordinates of point of division are

X =(k1.x2 +k2.x1)/(k1+k2) And Y =(k1.y2 +k2.y1)/(k1+k2)

Step-by-step explanation:

We can use the Ratio-Formula to find the co-ordinates of point of division.

If a point P divides a line segment AB with A(x1, y1) and B(x2, y2) in a ratio k1: k2, then the co-ordinates of the point of division will be

X =(k1.x2 +k2.x1)/(k1+k2) And Y =(k1.y2 +k2.y1)/(k1+k2)

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Here's Your Answer :

Length = 5 cmBreadth = 4 cmHeight = 3 cm

=>Total Surface Area of A Cuboid

2 ( l × b + b × h + h × l )2 ( 5 × 4 + 4 × 3 + 3 × 5 )2 ( 20 + 12 + 15 )2 × 4794cm

So, the total surface area of cuboid is 94cm

Area = length×breadth×height5×4×360 cm square is the area

as the graph is cutting the x-axis at 2 pointsso here 2 zero will be there

x= 2 ,3

In a Rhombus the diagonals are perpendicular to each other meeting at the center O.

In the triangle BOC, ∠O = 90°. Sin ∠OBC = OC / BC = (1/2 AC) / BC = 1/2 so ∠OBC = 30° so angle∠ABC = 2 * 30° =60°

Then ∠OCB-∠OBC = 60° so ∠DCB = 2 * 60° =120°

The angles of Rhombus are 60 and 120 deg.

4×4=16,6×6=36,8×8=64 are perfect squares

16 , 36 , 64are the perfect squares numbers

16 is square of 436 is of 664 is of 8

[5^n*5^2 -6*5^n*5]/[13*5^n*1/5^2 -2*5^n*5]

=[5^n(25-30)]/[5^n(13/25 -10)]after cancellation=(-5)/[(13-250)/25]

=(-5)/[(-237)/25]=(5*25)/237=125/237

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14+15÷21 29÷21=1.4

29/21

2/3+5/7

=14+15/21=29/21 is answer.

Given :- Length of the chord AB = 16cm and radius ( r ) = 10cm .

To find :- OM

Construction :- Draw a perpendicularOM from centre O such that, it bisects the chord AB.

Proof :- In right angled ∆OMB,

OB^2 = OM^2 + BM^2

=> 10^2 = OM^2 + 8^2

=> 100 = OM^2 + 64

=> OM^2 = 36

=> OM = √36

=> OM = 6 cm

16

square of 16 is 4 , squre of 36 is 6, 64=8

Red line is x = -6.Blue line is y = 1.

26 is the correct answer of the given question is

19+33÷252÷2=26 26 is right answer

(19÷2)+(33÷2)= (19+33)÷2= 52÷2 = 26 is the best answer

Given:

Height (h) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 3 m

Radius of the cylindrical part = 3/2 m

Slant height (l) of conical part = 2.8 m

Total canvas used = CSA of conical part + CSA of cylindrical part

= πrl + 2πrh

= πr(2h+l)

= (22/7)×3/2(2×2.1+2.8)

= (22/7)×3/2(4.2+2.8)

= (22/7)×3/2(7)

= 11×3

= 33 m²

Cost of 1 m² canvas = ₹ 500

Cost of 44 m² canvas = 33 × 500 = 16500

The cost of Canvas needed to make the tent= ₹ 16500

Hence, it will cost ₹ 16500 for making such a tent.

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A = {3,7,9,10,13,15} B = {2,3,5,7,11,13,17,19} B-A = {2,5,11,17,19}

-(-x)=xfor x=2/15-[-(2/15)]=-1×-1×2/15=2/15 = x

for x=-13/17=-[-(-13/17)]=-[-1×-1×13/17]=-13/17 = x

*In the first part of the solution, in matrix B, the number is 30. It's not 3.*

According to question,(2z - x - 3y) - ( x - y + 3z)

2z - x - 3y - x + y -3z

(2z - 3z) + ( -x -x) + (-3y + y)

- z - 2x - 2y

We can obtain this by subtracting both2x-4y-z-(x-2y+3z)=x-2y+2z

Principal amount= 18000 ruppesTime = 2 yearsrate of interest= 10A= P(1+R/100)^2= 18000(1+10/100)^218000(110/100)^21.21*1800021780 ruppesThis will be the amountInterest = 21780-180003780 ruppes

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440000

try doing the method u think will be correct

A simple journal entry has adebitand credit of equal value. For example, a $12,000 business vehicle purchased withcashis recorded as a $12,000debitto equipment and a $12,000 credit tocash. A compound journal entry has multiple debits, multiple credits or both debits and credits.

if AC = BC ,

then AC = BC , should be equal to the altitude OC = 6.5cm

so, AB = 6.5+6.5 = 13cm

now area base √13²-12² =5cm , using Pythagoras theorem,

so, area ∆ AOB = 1/2*(12)*(5) = 30cm²

The maximum amount of deduction that can be claimed under section 80C is Rs 1.5 lakh for the current financial year. The section offers various investment options to the taxpayer which not only generates returns for him but can also be claimed as deduction while calculating total taxable income.

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Let P = 5 kg weight = 5 g NewtonsQ = 10 g Newtons

Using the parallelogram law of addition of vectors :Resultant forcemagnitude = R R² = P² + Q² + 2 P Q CosФ = 5² + 10² + 2 * 5 * 10 * Cos 90 = 125R = 5√5newtons

Angle made by R with Q = 10 kg weight =α tanα = P Sin Ф/ ( P + Q Cos Ф) = 5 Sin 90 / (10 + 5 Cos 90) = 5/10=1/2

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x is equal to X y is equal to a b a b x d c is equal to DC is equal to ab AC by BC is equal to y ABCD

Given,P = 60000, R = 9%, n = 2

For compound interestA = P(1 + R/100)^nA = 60000(1 + 9/100)^2A = 60000(1.09)^2A = 60000(1.1881)A = 71286

Therefore Rohit will receive amount of Rs 71286 at the end of 2 years.