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Step-by-step explanation:

Worker paid rupees = 309

Days = 6

By proportional method

309:6::978.50:x

Let No be x

309x=978.5

978.5/309 =x

19= x

B) no. Of days in January =31

309/6 = one day

309/6*31

1596.5 Ans

(root(3)+root(2))(root(3)-root(2))=3-2=1

(√3)^2- (√2)^2 = 3 - 2 =1

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500 notebooks required paper = 45 kg 1 notebook required paper = 45/500 kg700 notebooks required paper= (45/500)*700= 9*7= 63 kg

વિકલ્પ બી બરાબર છે, આ 11 અને 12 ની વચ્ચે 80/7 = 11.42

Arithmetic mean = sum of all values/total number(7 + 8 + x + 11 + 14)/5 = x 40 + x = 5x 4x = 40 ∴ x = 10

Arithematic mean will be 20+11+21+25+23+14/6=19

Let a and d be the first term and common difference of the AP.

given, pth term of the AP = q and qth term of the AP = p.

To show: (p+q)th term is zero.

a + (q-1)d = p ------- (1)a + (p-1)d = q ------- (2)

Subtract (1) from (2)

(p-1)d-(q-1)d = q-p=> (p-q)d = q-p => d = -1 ----- (3)

Using (1) and (3) we get,

a + (q-1)(-1) = p=> a = p+q-1 -----(4)

(p+q)th term = a + (p+q-1)(d) = (p+q-1) + (p+q-1)(-1) {from eq. 3 and 4}(p+q)th term = 0 Hence showed.

Arithmetic mean = sum of all the number/total number5 = sum of all number/7sum of all number = 5*7 = 35

Inner circumference = 220

2 π r = 220

r *2* 22 = 220 * 7

r = 35 m [ inner radius ]

inner radius , r₁ = 35 m outer radius , r₂ = 35 + 7 = 42 m

outer circumference = 2 π r ₂

= 2 * ( 22 / 7 ) * 42 = 264 m

Cost of fencing = Rs 2 * 264 = Rs 528

ad=epangle epa=dpb(given)angle bad=abe(given)by AAs both the triangle are congruentit meansdap congruent to ebpso,ad=be (by cpct)

Inner circumference = 220

2 π r = 220

r *2* 22 = 220 * 7

r = 35 m [ inner radius ]

inner radius , r₁ = 35 m outer radius , r₂ = 35 + 7 = 42 m

outer circumference = 2 π r ₂

= 2 * ( 22 / 7 ) * 42 = 264 m

Cost of fencing = Rs 2 * 264 = Rs 528

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i) R1 is not a function as for the value of 2 , we have , two values 3 and 7. but the condition is for 1 element in x element , it should be related to only one in y.

ii) this is a function.. because for every every value of x, we will get different value of y.

A = { x² | x is natural numbers}

Place value =600Face value =6Diffrence =600-6=594

ogive gives cumulative frequencyit gives upperclass limit hence it gives median

Let b=na. Since b/a is an integer and both b and a are positive, n is also a positive integer.

Since the numbers are in geometric progression, it follows that:c=n^2 a

Given that their arithmetic mean is b+2 (=na+2), it means:(a + na + n^2 a)/3 = na+2

Simplifying,n^2 -2n + (1 - 6/a) = 0

Finding the roots of this quadratic equation in n,n =1 ± 0.5√(24/a) [using quadratic formula]Since n is an integer, 24/a has to be a perfect square. Hence the only possible value of a is 6, which gives n=2

So the values of a, b, c are 6, 12, 24, respectively.Now putting the values we get,a2+ a -14/a+1=6^2+6-14/7=4

put x=0. 2x-3y=123y=6 . put x =0y=6/3. -3y=12y=2. y=12/-3points (0,2) . y=-4 x+3y=6. points(0,-4)put y=1. put y=0x+3=6. 2x=12x=6/2. x=12/2x=3. X=6points(3,1). Points(6,0)

4x+3y=30x=(30-3y)/4y=2,x=6y=6,x=3y=10,x=0y=14,x=-3y=18,x=-6

thanks di

1) 9004) 22this is the correct answer for this question Thank you

9(i) √625×√129625×32= 800

800 is the right answer

9(i) 800 is the right answer

900 is write answers

4.)√2116=46√1764=4246-42/46+42=4/88=1/22

1.)√625=25√1296=32

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25-15×15+4+200204-200=4

i) Sum to n terms: S₁ = a*n(n - 1) ------ (1) [Given]So sum to (n - 1) terms of AP is: S₂ = a*(n - 1){(n - 1) - 1)} = a(n - 1)(n - 2) ----- (2)

Hence nth term of AP = S₁ - S₂ [Eq (1) - Eq(2)]:

= a*n(n - 1) - a(n - 1)(n - 2) = a(n - 1)(n - n + 2) = 2a(n - 1)

Substituting n = 1, 2, 3, 4, -------the AP is: 0, 2a, 4a, 6a, 8a, 10a, -----------

ii) Hence of the above, the sequence whose terms are square of each term:

0, 4a², 16a², 36a², 64a², ----

Sum of these squares is: 0 + 4a² + 16a² + 36a² + 64a² + ---

S = 4a²[0 + 1 + 4 + 9 + 16 + ---]

= 4a²[0² + 1² + 2² + 3² + 4² + 5² + ---]

= 4a²[1² + 2² + 3² + 4² + ------ + (n - 1)²]{Number of terms are only (n - 1) terms}

Sum of squares of first n natural numbers = n(n + 1)(2n + 1)/6Hence sum to (n - 1) terms = (n - 1)(n)(2n - 1)/6

Thus sum of the squares S = 4a²(n)(n - 1)(2n - 1)/6

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hi friend can you write it in English

thank you

log5(sqrt[625])

Rewrite as anequation.

log5(sqrt[625]) = x

Rewritelog5(sqrt[625] )=xin exponential form using the definition of a logarithm. Ifx andb are positive real numbers andb does not equal1, thenlogb(x)=y isequivalenttob^y=x

5^x = 25

Createequivalentexpressionsin theequationthat all have equalbases.

5^x = 5^2

Since thebasesare the same, the twoexpressionsare only equal if theexponentsare also equal.

x = 2

Thevariablex is equal to 2

2 log .base 5 and 5 square ,it's ans are 2

(4x+5)(2x-1) is the correct answer of the given question

The right answer is (4x+5)(2x-1)

(sinθ- 2 sin^3θ)/(2cos^3θ-cosθ)

=sinθ(1-2sin^2θ)/cosθ(2cos^2 θ-1)

= sin θ [1- 2(1- cos^2 θ)/cos θ(2cos ^2 θ -1)

= sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1)

= sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1)

= sin θ/ cos θ

= tan theta