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for general quadratic eqⁿAx²+Bx+C=0if roots are equal then√(B²-4AC)=0i.e.B²=4AC--------(1)here eqⁿ Is bx²+ax+c=0B=aA=bC=cput in eqⁿ1a²=4bc

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Chbilalakbar

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Answer:First natural number is 10 and second is 20Explanation:Let x and y be natural numbers such that x is less then y

According to given conditionAverage of x and y = (x + y)/2 = x + 5 ..... (1)Ratio of numbers = y/x = 2 ..... (2)⇒ y=2x .....(3) using equation (3) in (1) we get (x + 2x)/2 = x + 5

Multiplying by 2 on both sides

x+2x=2x+10

Subtracting 2x from both sides

x=10and using x=10 in equation (3) we get y=2(10)=20Hence First natural number is 10 and second is 20

Let x and y be natural numbers such that x is less then y

average of x and y = (x + y)/2 = x + 5 ..... (1)

ER. RAVI KUMAR ROY

Ratio of numbers = y/x = 2 ..... (2)

⇒ y=2x .....(3)using equation (3) in (1) we get

(x + 2x)/2 = x + 5

Multiplying by 2 on both sides

x+2x=2x+10

Subtracting 2x from both sides

x=10

and using x=10 in equation (3) we get

y=2(10)=20

Hence First natural number is 10 and second is 20..

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eya copy me bane ga bro isme nhi Bana sakte hai number line

surface area of cube 6×a², is a side of cube 6a² = 600a² = 100a = 10

Please post the picture of the equation. It is hard to understand the question

Let the radius of a circle be r.

∴ Circumference of a circle = 2πr

Let the radii of two circles are r and r whose values are 15 cm and 18 cm respectively.

i.e. r1 = 15 cm and r2 = 18 cm

Now, by given condition,

Circumference of circle = Circumference of first circle + Circumference of second circle

⇒ 2πr = 2πr1 + 2πr2

⇒ r = r1 + r2

⇒ r = 15 + 18

∴ r = 33 cm

Hence, the required radius of a circle is 33 cm.

GIVEN: p(x) = ax^4+2x^3–3x²+bx

g(x) = x²-4

Where p(x) & g(x) are polynomials in variable ‘x'

& here g(x) is a factor of p(x)

That means p(x) is exactly divisible by g(x)

Or p(x) is exactly divisible by factors of g(x)

Now, factors of g(x) = x²-4 = (x+2)(x-2)

So, p(x) is exactly divisible by (x+2) & (x-2)

That means, if p(x) is divided by (x+2) & then by (x-2) , the remainder has to be zero.

So now we find out the remainder in each case by remainder theorem:

If p(x) ÷(x+2) , the ramainder = p(-2)

ie, p(x)= ax^4+2x^3–3x²+bx is divided by (x+2)

Remainder= p(-2)= 16a - 16 -12 -2b =0………(1)

& if p(x)= ax^4+2x^3–3x²+bx is divided by (x-2)

Remainder= p(2)= 16a+16–12+2b=0……….(2)

eq(1) +eq(2)

=> 32a-24=0

=>32a =24

So a= 24/32 = 3/4………(3)

Now, eq(1) _ eq(2)

=> -32 -4b =0

=> 4b = -32

=> b= -8…………(4)

ANS a= 3 /4

b= —8

=2√2+5√3+√2-3√3=3√2-2√3

Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r which satisfies the condition

a = bq + r

where 0 ≤ r < b .

Let the first number be xSecond be x+1 and third be x+2

According to the question :(X+1+X+2)-X=142X+3-X=14X=14-3X=11

THEREFORE THE NUMBERS ARE 11,12 AND 13

can u pls click the photo clear

7x^2-x(x-2)=7x^2-x^2-2x=6x^2-2x ;; (iii)2x(x+5)+1=2x^2+10x+1 ; (iii)2x-1; 2x=1; x=1/2

x+3/x=x^2:; x+3=x^3; x^3-x=3=x^2

thanks for answering

The answer is =810,000 of 30000×9=270,000×3=810,000

Amount = P(1 + (r/100))^n = 30000(1 + (9/100))^3 = 30000(109/100)^3 = 3 × 1295029/100 = 38850.87Compound Interest = A - P = 38850.87 - 30000 = 8850.87

Let the two nos be x and y

Now given -: Difference of two numbers which are perfect cubes is 189

⇒x^3 - y^3 = 189

Let y be the smaller of the two nos

and the cube root of the smaller [y] of the two numbers is 3

⇒ y=3

find the cube of 3 and that is 27

⇒ y^3 = 27

ATP

⇒ x^3 - y^3 = 189

=> x^3 -27 =189 [Putting the value of y^3 = 27]

=> x^3 =189 +27

=> x^3= 216

Cube root of (216) = 6

⇒ x=6

Therefore, the larger number is 6. Ans

and the smaller number is 3 Ans

Let's, first thing we should do is to find the cube of 3 and that is 27. Now, add 27 to 189 - that is 27 + 189 = 216 Finally, find the cube root of 216 and that is 6. Therefore, the cube root of the larger number is 6.

is the correct answer

6 is the right answer of this question. please like my answer

number 6is the correct answer

The cost of larger box is Rs 197.6 and smaller box is Rs 63.232

Step-by-step explanation:

i) Bigger box:

L = 20 cm = 0.2 m

W = 15 cm = 0.15 m

H = 5 cm = 0.05 m

hence total surface area = 2(LW + LH + WH)

= 2(0.2x0.15 + 0.2x0.05 + 0.15x0.05)

= 2(0.03 + 0.01 + 0.0075)

= 0.095 m²

4% of TSA is used for overlapping = 0.04 x 0.095 = 0.0038

Hence Net card board used for 1 larger box = 0.095 + 0.0038 = 0.0988

Net card board used for 200 larger box = 200 x 0.0988 = 19.76 m²

Hence total cost of 200 box Rs 10 per m² = 10 x 19.76 = 197.6

ii) Smaller box

L = 16 cm = 0.16 m

W = 12 cm = 0.12 m

H = 4 cm = 0.04 m

hence total surface area = 2(LW + LH + WH)

= 2(0.16x0.12 + 0.16x0.04 + 0.12x0.04)

= 2(0.0192 + 0.0064 + 0.0048)

= 0.0304 m²

4% of TSA is used for overlapping = 0.04 x 0.0304 = 0.001216

Hence Net card board used for 1 smaller box = 0.0304 + 0.001216

= 0.031616

Net card board used for 200 smaller box = 200 x 0.031616 = 6.3232 m²

Hence total cost of 200 box Rs 10 per m² = 10 x 6.3232 = 63.232

Hence the cost of larger box is Rs 197.6 and smaller box is Rs 63.232

1. For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such thata=bq+r, where0≤r<bExplanation:Thus, for any pair of two positive integers a and b; the relationa=bq+r, where0≤r<bwill be true where q is some integer.

2. The fundamental theorem of arithmetic (FTA), also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is prime itself or is the product of a unique combination of prime numbers

Let one number is x and other is x+36Sum=48x+x+36=482x=12x=6

Numbers are 6 and 42

excellent

The smallest four digit number is, obviously, 1,000.

1,000 is a multiple of four (because it’s a multiple of 100.)

So the biggest 3-digit multiple of 4 must be 1000 - 4

that’s to say, 996.

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A)12% of x=6012x/100=60X=60*100/12=500B)25% of x=7025x/100=70X=70*100/25=280C)65% of x=22165x/100=221X=221*100/65=340D)12.5% of x=100012.5x/100=1000X=1000*100/12.5=8000

Answer:We can estimate the cube root by the splitting the number from the right into three digit numbersSo for 1331Left group 1Right group 331As you know 13= 1 so there would be 1 at unit’s place in cube root of 1331.Now we have to find the cube root of left group 1Now 13= 1So, we have 1 in ten’s place and 1 in unit place113= 1331 satisfies the condition4913:Right group = 913Left group = 473gives 3 at unit’s place so unit digit number in cube root of 4913 should be 7We have to estimate the cube root of left group i.e 413= 1 and 23= 81<4<8So, 10s digit in cube root of 4913 should be 1So answer is 17

The numbers between 1 to 1000 which are multiple of 60(LCM(2,3,4,5))are divisible by 2,3,4,5.

They are 60,120,180,….960

This is an A.P

an=a1+(n−1)dan=a1+(n−1)d

960=60+(n−1)60960=60+(n−1)60

900=(n−1)60900=(n−1)60

n−1=15n−1=15

n=16n=16

Answer is 16

Thnks

Let the numbers be a-d,a,a+dHence sum is 18 that isa-d+a+a+d=3a=18a=6Now product is 162Hence(a-d)*a*(a+d)=162(6-d)*6*(6+d)=162(36-d^2)*6=16236-d^2=27d^2=9d=+-3If d=3the ap will be 3,6,9,..If d=-3then ap will be 3,0,-3,-6,..

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Volume of cuboid = LBH = 7×8×9 = 504Volume of cube = (5) ³ = 125 Volume of remaining cuboid = = 504-125 = 379

1. A straight line segment can be formed by joining any two points in space.

2. All right angles are congruent or equal to one another.

1408 is the correct answer.

837+208+363=1408 ans

Given summation is 837 + 208 + 363 = 1408.

total answer is 1408