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cost of 1kg sugar=1750/50=35 rs /kgcp=35rs/kgsp=32 rs/kgloss=cp-sp=35-33=3total less money get =50*3=150 rsorsp of 50kg sugar=50*32=1600cp= 1750loss=1750-1600=rs150he get rs 150 less than he had spend

thanks

it is incomplete

but thankyou gor help

If contains 60 dm 3 volume of watervolume = l×b×h60,000= 50×40×hh= 600/2030 cmThe height of the water in the fish tank is 30 cm

It contains 60 dm^3 volume of water. 60 dm^3=60000 cm^3Volume= l×b×h60,000=50×40×h h=600/20 =30 cm The height of the water in the fish tank is 30 cm

1/2 ( sum of d parallel sides) × height= 1/2×122×h=1586h=26cm

h= 22 cm is the height of trapezium

Cos(40+x)-Sin(50-x)+ Cos^2 40+ cos^2 50/ Sin^2 40+ Sin^2 50=Cos(40-90)-Sin(40-90) + cos^2(40+50)+ Cos^2(50+40)/Sin^2(40+50)+Sin(40+50)= Sin50-Sin50+ cos^2(90)+Sin^2(90)/ Sin^2 (90)+ Sin^2(90)=1-1+1/1=0

cos(40+90)- sin(50-90)+ cos^40+ cos^2(90-50)/sin^2(90-40)+ sin^2(50)= cos^2 (40)+ sin^2(40)+ sin^2(40)+ sin^2(50)/ cos^2(50)+ cos^2(50)=1+1×1

1 is the correct answer of the given question.

25a^2 + 40ab + 16b^2

= (5a)^2 + 2*5*4*ab + (4b)^2

[using (a + b)^2 = a^2 + b^2 + 2ab]

= (5a + 4b)^2

= (5a + 4b)(5a + 4b)

Enter a problem...

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Algebra

Factor 25a^2-40ab+16b^2

25a2−40ab+16b225a2-40ab+16b2

Rewrite25a225a2as(5a)2(5a)2.

(5a)2−40ab+16b2(5a)2-40ab+16b2

Rewrite16b216b2as(4b)2(4b)2.

(5a)2−40ab+(4b)2(5a)2-40ab+(4b)2

Check the middletermby multiplying2ab2aband compare this result with the middletermin the originalexpression.

2ab=2⋅(5a)⋅(−4b)2ab=2⋅(5a)⋅(-4b)

Simplify.

2ab=−40ab2ab=-40ab

Factorusing theperfect squaretrinomialrulea2−2ab+b2=(a−b)2a2-2ab+b2=(a-b)2, wherea=5aa=5aandb=−4bb=-4b.

(5a−4b)2

25a^2 +40ab+16b^2

=(5a)^2+2*5*4*ab+(4b)^2

[using(a+b)^2=a^2+b^2+2ab]

=(5a+4b)^2

=(5a+4b)(5a+4b)

Kindly reject 1st answer

15/75=1/540/90=4/9hence1/5+4/99+20/45=29/45

In order to find rational numbers between 3/5 and 7/8take LCM of 5 and 8

Now, find rational number between3/5 = 24/407/8 = 35/40

Therefore,Three rational numbers 24/40 and 35/40 are 25/40, 26/40 and 27/40

Infinite rational numbers can be found between 3/5 and 7/8

happy environment day

To find three rational numbers lying between 3/5 and 7/8

Take LCM of 5 and 8 = 40

3/5 = 24/407/8 = 35/40

Three rational numbers between 24/40 and 35/40 are25/40, 26/40, 27/40

There are infinite rational numbers between these two numbers.

-7/2 and -2-7/2=-3.5hence-3,-2.5,-2.75

6/7 & 5/9 l.c.m of 7 and 9 = 637 *9=63 6*9= 54 =54/639*7=63 5*7=35=35/63now we need to find the nos. so preferably we shouldmulyiply the fractions with numbers like 10,100,1000 etc.54*10=54063*10=63035*10=35063*10=630numbers between 540 and 350

351/630 , 352/630 , 353/ 630

To find three rational numbers between 2/3 and 3/4

Multiply and divide first rational number by 16 and second rational number by 12

Then,2/3 = 32/483/4 = 36/48

Therefore,Three rational numbers between 32/48 and 36/48are 33/48, 34/48, 35/48

6=2×3

Consider the highest factor which is 3.

Divide 125 by 3 = 41.66

Take its greatest integer =41.

Now divide 41 by 3= 13.66.

Take its greatest integer =13.

[13/3]=4.

[4/3]=1 [1/3]=0

Now add 41+13+4+1+0=59

(1). Given that, D =-5.5then focal length =1/D100/-5.5 (1m=100cm)-18.18cm is the focal length of lens with -5.5 power(2)Given that, D=1.5then focal length =1/D

Given that D =-5.5then focal length =1/D100/-5.5 (1m =100cm) -18.18cm is the focal length of eye lens with -5.5 DDistant vision=f=-DDistant vision= 18.18cm(2)given D =1.5then focal length =1/D 100/1.5=66.6cmNear vision

(i)For distant visionPower of lens=-5.5 Dtherefore focal length=1/-5.5 m=(1x10)/55 m=0.18m(ii) for near visionpower=+1.5 Dtherefore focal length=1/1.5m=(1x10)/15m=0.66m

Total of 11 observations 60 × 11 = 660 Total of first six observations 6 × 58 = 348 Total of last six observations = 61 × 6 = 366 So 6th observation 348 + 366 - 660 = 54

The center of a circle passing through 3 non-collinear points is at the intersection of the perpendicular bisectors of the segments connecting those points.

Sketch the figure of three such points, connect them with segments, and sketch each perpendicular bisector. Then sketch a segment from each point to the point of intersection of the bisectors. Using the SAS postulate you can prove the triangles formed by the bisector of one of these segments, half of the bisected segment and the segment you added are congruent triangles, and because corresponding parts are therefore congruent, and using the transitive property of congruence, you can prove the distances from the three points to the intersection point ate congruent. Thus that distance represents the radius of a circle connecting the three points.

Arrange the six numbers in order, 3, 5, 5, 7, 8, 9

We have three more numbers to insert into the list, and the median will be thehighest 5(and5lowest) number on the list. If the three numbers aregreater than9, the median will be the highest it can possibly be. Thus, the median is the 5th item of data which is8

1

2

Surface area of sphere = 4 × pi × r²

i) 4 × 22/7 × (10.5)² = 88/7 × (10.5)²

= 1386 cm²

ii) 4 × 22/7 × (5.6)² = 88/7 × (5.6)²

= 394.24

Meanscore of 9 innings is 58total runs in 9 innings = 58*9 = 522runs should scored in 10th innings to mean score 61 be X

61=522+X/10610=522+XX= 88

If 5 ounce peas cost $0.85Then 1 ounce of peas cost= 0.85/5 = $0.17

If 11 ounce peas cost $2.20Then 1 ounce of peas cost= 2.20/11 = $0.20

Therefore, 5 ounce peas for $0.85 is better buy

Let the Water tap B takes the time to fill the tank = y minutes

Now according o 1st condition

The Water tap A takes the time to fill the tank = (y + 7) minutes

Also according to second condition

Time taken by Tap A + 16 = Time taken by Both the taps

y + 7 + 16 = y + ( y + 7 )

y + 23 = 2 y + 7

Rearranging the above equation, we get

2 y + 7 = y + 23

2 y - y = 23 - 7

y = 16

So Water tap B takes the time to fill the tank = 16 minutes

And

Water tap A takes the time to fill the tank = 16 + 7 = 23 minutes

So Tap B alone takes 16 minutes and Tap A alone takes 23 minutes to fill the tank.

srry but the answer given in the answer sheet is this

Let the time taken by A is x minutes and B is y minutes,

Hence from the 1st statement we have,

x-y = 7 = > y = x-7..........................eq1

Now, in one minute tank filled by A = 1/x

and in one minute tank filled by B = 1/y

So combine together both can fill = 1/x + 1/y tank in one minute,

Hence time taken by both the taps to fill the tank

= 1/(1/x + 1/y)

= xy/x+y

Now from the second statement,

x- xy/x+y = 16

=> x² + xy - xy = 16(x+y)

=> x² -16x - 16y = 0

Putting the value of y from eq1

x²-16x - 16(x-7) = 0

=> x² - 32x + 112 = 0.

Solving the above quadratic eqn, we get

x = 28 or 4

rejecting 4 as x can't be less than 7

x = 28

y = x-7 = 28-7 = 21

Hence time taken by both the taps are 28 and 21 minutes alone.

Let Tap B takes x mins

Tap B takes x + 7 mins

Tap B takes x mins

⇒ 1 min = 1/x of the tank filled

Tap A takes (x + 7) mins

⇒ 1 min = 1/(x + 7) of the tank filled

Together:

1 min = 1/x + 1/(x + 7)

1 min = [ x + 7 + x] /x(x + 7)

1 min = (2x + 7) /x(x + 7) of the tank

Total time needed to fill the tank together:

mins needed = x(x + 7)/(2x + 7)

The tap A takes 16 minutes more than the time taken by both the taps together to fill the tank:

Tap A = Tap A and Tap B + 16 mins

(x + 7) = x(x + 7)/(2x + 7) + 16

x + 7 - 16 = x(x + 7)/(2x + 7)

x - 9 = x(x + 7)/(2x + 7)

x(x + 7) = (x - 9)(2x + 7)

x² + 7x = 2x² + 7x - 18x - 63

x² - 18x - 63 = 0

(x - 21)(x + 3) = 0

x = 21 or x = - 3 (rejected, time cannot be negative)

Find the time needed:

Tap B = x = 21 mins

Tap A = x + 7 = 21 + 7 = 28 mins

Answer: Tap A takes 28 mins and Tap B takes 21 mins

Outer dimensions = L = 10 cm, B = 9 cm and H = 2.5 cmOuter volume = L×B×H = 10 × 9 × 2.5 = 225 cu cm

Thickness = 0.5 cm

Inner dimensions =l = L - 2w = 10 - 2(0.5) = 9 cmb = B - 2w = 9 - 1 = 8 cmh = H - w = 2.5 - 0.5 = 2 cm (Open at the top)

Inner volume = lbh = 9 × 8 × 2 = 144 cu cm

Volume of metal = Outer volume - Inner volume = 225 - 144 = 81 cu cm

I am understand questions

(i)Radius (r) = 7 cm

Slant height (l) = 25 cm

Let h be the height of the conical vessel.Slant height (l)²=r²+h^2

h = √l² – r²h = √25²– 7² = 625- 49h = √576h = 24 cm

Volume of the cone = 1/3 πr²h

= (1/3 × 22/7 × 7 × 7 × 24)= 1232 cm³[1 cm³= 1/1000L]Capacity of the vessel = (1232/1000) L= 1.232 LCapacity of thevessel =1.232 L

(ii)

Given: Height(h) = 12 cm

Slant height (l) = 13 cmLet r be the radius of the conical vessel.

Slant height(l)²= r²+h²r = √ r² - h²r = √13²– 12²= √169 – 144r = √25r = 5 cmVolume of the cone = 1/3 πr²h= (1/3 × 22/7 × 5 × 5 × 12) = (2200/7) cm³Capacity of the vessel = (2200/7× 1000) L

= 11/35 lCapacity of the vessel =11/35 L