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Price × consumption = constant

initial price = pchanged price = p' = 1.2pinitial consumption= Cchanged consumption= c'

p'c' = pc

1.2p × c' = pc

c' = pc/1.2p

c' = c/1.2 = 0.83c

Reduction in consumption is 17%

I cant understand

Right answer is

Let the original Price of Sugar = x Rs/kg

Let The Original Consumption = y Kgs

Hence Original Expenditure = Rs. (xy)

New Price = x + (20/100)x = 1.2x

Let the new consumption = z

New Expenditure = 1.2xz

But according to question

xy = 1.2xz

⇒ y = 1.2z

⇒ z = y/1.2 = (5y/6)

Let the Percentage Decrease in Consumption = p

y - (p/100)y = 5y/6

⇒ 1 - (p/100) = 5/6

⇒ 1/6 = p/100

⇒ p = 100 / 6 = 50 / 3 = 16.66%

√2x² + 7x + 5√2 = 0

√2x² + 2x + 5x + 5√2 = 0

√2x ( x + √2) + 5 ( x + √2) = 0

(√2x + 5)(x + √5) = 0

So, x = -5/√2 and x = -√5

Volume of the godown = 240x 210 x 160 = 8064000m^3Therefore no: of cuboildalboxes of 0.8 m^3 thatcan fit in the godown = 48000/0.8=10080000

x2+17=0x2=-17x= root-17

Let mid point of A and B is PP=(0+2)/2,(-1+1)/2=(1,0)Let mid point of B and C is QQ=(0+2)/2,(3+1)/2=(1,2)Let mid point of C and A is RR=(0+0)/2,(-1+3)/2=(0,1)Now area of triangle PQR(1/2)[1(2-1)+1(1-0)+0(0-2)](1/2)[1+1+0]=1sq. unit

(x^2 - 3x + 4) + (4x^2 - 2x + 7) + (-x^2 + 5x - 2)= (1+4-1)x^2 + (-3-2+5)x + (4+7-2)= 4x^2 + 0*x + 9= 4x^2 + 9

write

worg

8x-3=9-2x8x+2x=9+310x=12x=12/10=6/5

The value is 5^-2=1/5^2 = 1/25

The minimum calorie intake decided by the planning commission for rural and urban areas are 2400 and 2100 respectively.

Out of 452/5 are badthat means 2/5*4518 are bador(45-18)are good27 are good

given equation is=2x+3y=6---(1)

we rewrite the equation ,=3y=-2x+6

y=-(2/3)x+2

in these equation,slope=-2/3

the coordinating point here is (o,2)

set y=0 then ,0=-(2/3)x+2

2/3x=2

x=3

here the coordinating point is,(3,0)

using this coordinate we draw one graph,then

from the graph height will be come,2

and length will be come,3

area =2*3/2=3

if a line is of the form x/a +y/b = 1 then x intercept is a , and y intercept is b.

so here x intercept = 3 and , y intercept = 4

AB & AC are coordinate axes. BC line represents 3x + 4y =24For x=0, y= 6& for y=0, x= 8Hence, the vertices of right triangle are A(0,0), B(8,0) & C(0,6)So, AB =8, AC = 6 & BC = 10 ( by Pythagoras law)And ADOF will be square with side r.Inscribed circle is with centre O & radius rSince inscribed circle centre lies on the angle bisectors of the triangle.So, triangle ODB is congruent to triangle OEB (by RHS congruence criterion, or by AAS congruence corollary)So, BD = BE = (8-r)Similarly CF = CE = (6-r)Therefore, BE + CE = (8-r) +(6-r)=> BC = 14 -2r=> 10 = 14–2r=> 2r = 4=> r= 2

mujhe nahi aata ha bhai

the first answer is 0

11⅓ is the right answer

11 _________3) 34 quotient = 11 - 3 remainder = 1 ______ 04 - 3 ---------- 01

Price × consumption = constant

initial price = pchanged price = p' = 1.2pinitial consumption= Cchanged consumption= c'

p'c' = pc

1.2p × c' = pc

c' = pc/1.2p

c' = c/1.2 = 0.83c

Reduction in consumption is 17%please like the solution 👍 ✔️

Its 0,0 for originx= 0 y= 0

(x, y)=(0,0) is the coordinate of origib of coordinate axes

6x³ × 5x² = 30x⁵

Multiply the numbers and write the variable with its powers added.

the value of (6x^3)(5x^3) is 30x^5

210 = 55 * 3 + 4555 = 45 * 1 + 1045 = 10 * 4 + 510 = 5 * 2

hence 5 is the HCF.

5 = 45 - 10 * 4 substitute the values of 10 and 45 from above equations. = 210 - 55 * 3 - (55 - 45 ) 4 = 210 - 55 * 3 - 55 * 4 + 45 * 4 = 210 - 55 * 7 + (210 - 55 * 3) * 4 = 210 * 5 - 55 * 19

so x = 5 and y = -19

Let us first find the HCF of 210 and 55.

Applying Euclid division lemna on 210 and 55, we get

210 = 55 × 3 + 45 ....(1)

Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get

55 = 45 × 1 + 10 .... (2)

Again, considering the divisor 45 and remainder 10 and applying division lemna, we get

45 = 4 × 10 + 5 .... (3)

We now, consider the divisor 10 and remainder 5 and applying division lemna to get

10 = 5 × 2 + 0 .... (4)

We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55.

∴ 5 = 210 × 5 + 55y

⇒ 55y = 5 - 1050 = -1045

⇒ y = -19

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Let us first find the HCF of 210 and 55.

Applying Euclid division lemna on 210 and 55, we get

210 = 55 × 3 + 45

55 = 45 × 1 + 10

45 = 4 × 10 + 5

10 = 5 × 2 + 0

We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55.

∴ 5 = 210 × 5 + 55y

⇒ 55y = 5 - 1050 = -1045

∴y = -19

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TheCommunal Awardwasannouncedby theBritish Prime Minister, Ramsay MacDonald, in August1932.

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Tq

Answer =2775.8 cm square

D is the correct answer

base( b)=7cmheight (h)=3cmarea of parallelogram=base × height=>b×h=>7cm×3cm=21cm^2

We know that every quadratic equation is based on this relation..

p(x) = kx² - (α+β)x +αβwhere, α and β are the zeroes of given polynomial.

Now putting values of (α+β)and αβ in above equation, we get ...

p (x) = x² - (√2)x + (-3/2) = 0x² -√2x - 3/2 = 0

Let us multiply both sides by 2, we get...

2x² - 2√2x -3 = 0

Hence, the required polynomial is 2x² - 2√2x -3.

escape velocity and that is 11.2km/ second

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