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Product of number of days and number of workers remain constant. => 8*24= 6*X=> X = (8*24)/6=> X = 32 days.

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If we insure a Principal amount of P at a compound interest at rate 'r' for a period of 'n' years, then principal along with interest would be equal to

= P(1+r)^n

But here, principal amount is insured every year for 8 years

Now, given r = 12%

For initial amt insured, P(1+r)^8

For amt insured after 1yr, P(1+r)^7

For amt insured after 2yr, P(1+r)^6

......so on till 8 yrs

So, total returns = P[(1+r)^8 + (1+r)^7 +.....+1 ] = 15000(given)

=>P[(1+r)^9-1]/r=15000 ( Sum to 9 terms of a G.P)

On simplification we get, P = 1017 Rs

Gaurav has total money = 45.50Money bought book for 24.88Money left with Gaurav = 45.50 - 24.88 = Rs 20.62

If the actual measurement was 176 and the experiment measure was 20:

Error:

176−20=156

This as a percentage of the true measure:

(156/176)×100%=88.63%

Loss is C.P-S.Pso loss is 1000 rs.Loss %=loss/C.p*100=1000/15000*100=6.66%

ok

40percent of 20040/100*20080 childrens are girls there

dy/dx ={sinx(5x^4 + sinx) - (x^5 - cosx)(cosx)}/sin^2x={5sinx x^4 - cosx x^5 +(sin^2x + cos^2x)}/sin^2x=(5sinx x^4 - cosx x^5 + 1)/sin^2 x

18 workers were paid 4950 so with same rate 8 workers must paid = 4950 × 8 / 18 = 2200 Rs.

Let the S.P = 100then C.P. = 25

Profit = 75

Profit% = (75/25) * 100 = 300%

1. Let x be the original unit price of the item in rupees/kg

2. $240/x=the original amount or item bought in kg

3. The price of item increases by 25%; therefore, the new price of item is 1.25x

4. $240/1.25x=the amount of item that is bought in kg after the price increase

5. The problem states that after the price increase, the amount of item I buy decreases by 1.5kg. Therefore, the difference between the original amount (step #2) and the new amount (step #4) must be 1.5kg. So,

(240/x)-(240/1.25x)=1.5kg

6. Solving the above equation gives x=32. So the original price was 32 rupees/kg.

The new price is 32*1.25=40 rupees/kg

CP=45000loss=15%so SP=45000*0.85=38250

i) 4q(q+r) = 4q²+4qr

ii) ab(a-b) = a²b-ab²

Thanks sir

let y1=(sinx)cosx

taking log on both sides

logy1=cosxlogsinx

diff. w.r. to x

1/y1.dy1/dx=cosx.(1/sinx).cosx+logsinx.(-sinx)

dy1/dx=(sinx)cosx[cos2x- sin2x.logsinx]/sinx

lety2=(cosx)sinx

taking log on both sides

logy2=sinxlogcosx

diff.w.r.to x

(1/y2)dy2/dx=sinx.(1/cosx)(-sinx)+logcosx .cosx

dy2/dx=(cosx)sinx[-sin2x+cos2x. logcosx]/cosx

dy/dx= sinx)cosx[cos2x- sin2x.logsinx]/sinx

+=(cosx)sinx[- sin2x+cos2x. logcosx]/cosx

Total number of balloons = 2+3+4 = 9

Probability of red = 2/9

Probability of green = 4/9

Probability of blue = 3/9

a balloon seller had 9 balloons he sold 5balloons how many balloons are left

Number of mangoes sold in 4 days=249×4=996

Please have a look at the attachment for the solution of this question.

14x - 10 must be subtracted from p(x) 8x⁴ + 14x³ - 2x² + 7x - 8 so that the resulting polynomial is exactly divisible by 4x₂ + 3x - 2

CP = 1625 SP = 1850 PROFIT PERCENTAGE = (1850-1625)/1625 * 100 = 225/1625 * 100 = 13.85%

thanks

really better than scholar

As CP < SP, there is a profit,Profit = SP - CP = 600 - 500 = 100Profit % = (100/500) × 100 = 100/5 = 20%

thank you bro

add 800+900=1700answeris this

add 800 + 900 = 1700

add 800+900=1700this is the answer

800+900=1700your answer is 1700

add 800+900=1700 this is the correct answer

800+900=1700your answer is 1700

A transaction is a business event that has a monetary impact on an entity's financial statements, and is recorded as an entry in its accounting records. Examples of transactions are as follows: Paying a supplier for services rendered or goods delivered.

∫tan^3(x) dx from 0 to pi/4

∫tan(x)*tan²(x) dx from 0 to pi/4

∫tan(x)*(sec²(x)-1) dx from 0 to pi/4

=tan²(x)/2 + ln|cos(x)| evaluated from 0 to pi/4

=1/2 + ln(√2/2)

1 and 2 are polynomial in one variable 5 is a polynomial in three variable 3 and 4 are not

both 1and 2 are polynomials and have one variable

3 and 4 have no variable and polynomials

5 is a polynomials and its have one variable