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a=-4b=-3c= 2d= 5

is the ans

Use:Minus × Minus = PlusMinus × Plus = Minus

Minus ÷ Minus = PlusMinus ÷ Plus = Minus

a) (-400) × (-40) × (-4) = (-400) × 160= -64000

b)+186 ÷ (-31) = -6

Use:Minus × Minus = PlusMinus × Plus = Minus

Minus ÷ Minus = PlusMinus ÷ Plus = Minus

a) (-400) × (-40) × (-4) = (-400) × 160= -64000

b)+186 ÷ (-31) = -6

P(getting head ) = 100 / 200 = 1/ 2

(0.125/8 - 0.025/40)/(0.5×0.5×0.5 + 1/5) = (0.015625-0.000625)/(0.125+0.2) = 0.015/0.325 = 15/325 = 3/65

let no. be x...according to given question,40+5x=11x6x=40x=40/6

Explanation : 85 = 5×17 153 = 3×3×17 HCF is a common factor of both the numbers. HCF = 17 85m - 153 = 17 85m = 170 m = 170/85 m = 2

Answer : m=2

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As both have same base,

Equating LHS power

m - 2 + 3m = 13

4m = 15

m = 15/4

thank you

area is 1/2*b*hso altitude=(2*90)/20=9cm

Thank you so much

oil left = 18kg - 7 kg 50 gram = 18kg - 7. 05 kg = 10.95

492.5 L

You are genius

Books selected = 4C6= 4! /(6! ×2!) = 1/5×6×2= 1/60

15,360

Theorem:

Letf(x)be any polynomial of degree greater than or equal to one and let ‘a‘ be any number. Iff(x)is divided by the linear polynomial(x-a)then the remainder isf(a).

Remainder Theorem Proof:

Letf(x)be any polynomial with degree greater than or equal to1.

Further suppose that whenf(x)is divided by a linear polynomialp(x) = ( x -a), the quotient isq(x)and the remainder isr(x).

In other words ,f(x)andp(x)are two polynomials such that the degree off(x)is greater then equal todegree of p(x)andp(x)is not equal 0then we can find polynomialsq(x)andr(x)such that, wherer(x) = 0or degree of r(x) < degree of g(x).

By division algorithm

f(x) = p(x) . q(x) + r(x)

∴ f(x) = (x-a) . q(x) + r(x) [ here p(x) = x – a ]

Since degree of p(x) = (x-a) is 1 and degree of r(x) < degree of (x-a)

∴Degree of r(x) = 0

This implies that r(x) is a constant , say ‘ k ‘

So, for every real value of x, r(x) = k.

Therefore f(x) = ( x-a) . q(x) + k

If x = a,

then f(a) = (a-a) . q(a) + k = 0 + k = k

Hence the remainder whenf(x)is divided by the linear polynomial(x-a)isf(a).

Statement of Factor Theorem:

If f(x)is a polynomial of degreengreater and equal to 1and‘ a‘ is any real number then

(x -a)is a factor off(x),iff(a) = 0.

and its converse ” if(x-a)is a factor of a polynomialf(x)thenf(a) = 0“

Factor Theorem Proof:

Given thatf(x)is a polynomial of degreengreater and equal to1by reminder theorem.

f(x) = ( x-a) . q(x) + f(a) . . . . . . . . . .equation ‘A ‘

1 .Suppose f(a) = 0

then equation ‘A’->f(x) = ( x-a) . q(x) + 0 = ( x-a) . q(x)

Which shows that( x-a)is a factor off(x).Hence proved

2 .Conversely suppose that(x-a)is a factor off(x).

This implies that f(x) = ( x-a) . q(x) for some polynomial q(x).

∴f(a) = ( a-a) . q(a) = 0.

Hencef(a) = 0when(x-a)is a factor off(x).

The factor theorem simply say that If a polynomialf(x)is divided byp(x)leaves remainder zero thenp(x)is factor off(x

Example – 1 :Find the remainder when x3– 2x2+ 5x + 8 is divided by x + 1

Solution:Let f(x) = x3– 2x2+ 5x + 8

Now divisor = x + 1 , its zero is ‘ -1 ‘

By the remainder theorem, the required remainder = f( -1)

put x = -1 in above equation then we get

f (-1) = (-1)3– 2(-1)2+ 5(-1) + 8 = – 1 – 2 – 5 + 8 = 0

Remainder = 0

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do u really study in grade 8???

tq my friend

If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio. Let us now try to prove this statement.

Consider a triangleΔABC as shown in the given figure. In this triangle we draw a line PQ parallel to the side BC of ΔABC and intersecting the sides AB and AD in P and Q respectively.



Figure 1 Basic Proportionality Theorem

According the basic proportionality theorem as stated above, we need to prove:

ABPB=AQQC

Construction: Join the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the given figure.



Figure 2 Basic Proportionality Theorem- Proof

Now the area of ∆APQ =12× AP × QN (Since, area of a triangle=12× Base × Height)

Similarly, area of ∆PBQ=12× PB × QN

area of ∆APQ =12× AQ × PM

Also,area of ∆QCP =12× QC × PM ………… (1)

Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have

areaofΔAPQareaofΔPBQ=12×AP×QN12×PB×QN=APPB

Similarly,areaofΔAPQareaofΔQCP=12×AQ×PM12×QC×PM=AQQC………..(2)

According to the property of triangles, the triangles drawn between the sameparallel linesand on the same base have equal areas.

Therefore we can say that ∆PBQ and QCP have the same area.

area of ∆PBQ = area of ∆QCP …………..(3)

Therefore, from the equations (1), (2) and (3) we can say that,

ADPB=AQQC

Also ∆ABC and ∆APQ fulfill the conditions forsimilar trianglesas stated above. Thus, we can say that ∆ABC ~∆APQ

Yes the circle pass through point obecause o is the centre of ist circle and 2nd circle is drawn on the circumference of ist circle and both the circles have radius 5 cm

YES THEIR CORRESPONDING SIDES AND CORRESPONDING ANGLES ARE CONGRUENT

No value of k will satisfy both the equation simultaneously.

no from one point infinite lines can pass

2.only one line can pass through two distinct points

1)false2)falseThis is the answer

Replace P by C and solve.

Solution :

Let the two digit number be 10x + y where x is the tens digit and y is the ones digit.

Now, according to the question.10x + y = 8(x + y) - 510x + y = 8x + 8y - 510x - 8x + y - 8y = - 52x - 7y = - 5 .................(1)

And,10x + y = 16(x - y) + 310x + y = 16x - 16y + 310x - 16x + y + 16y = 3- 6x + 17y = 3 ................(2)

Now, multiplying the equation (1) by 17 and (2) by 7, we get

34x - 119y = - 85 ...............(3)-42x + 119y = 21 ..............(4)

Now, adding (3) and (4), we get

34x - 119y = - 85- 42x + 119y = 21_________________- 8x = - 64_________________

⇒ 8x = 64x = 64/8x = 8So, tens digit is 8.

Substituting the value of x = 8 in (1), we get2x - 7y = - 52×8 - 7y = - 516 - 7y = - 5- 7y = - 5 - 16- 7y = - 217y = 21y = 21/7y = 3

Ones digit is 3.So, the required number is 83.

thanks

thanks I got right answer