Mean of the following is 50 find p -Maths 9th

Mean of the following is 50 find p -Maths 9th
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2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

If the mean of the following data is 20.2, find the value of p: -Maths 9th

Description : If the mean of the following data is 20.2, find the value of p: -Maths 9th

Last Answer : x̅ = ∑fx/∑f ∴ 20.2 = 610 + 20p/30 + p ⇒ 20.2(30 + p) = 610 + 20p ⇒ 606 + 20.2p = 610 + 20p ⇒ 20.2p - 20p = 610 - 606 ⇒ 0.2p = 4 ⇒ p = 4/0.2 = 40/2 ⇒ p = 20

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Description : The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Last Answer : NEED ANSWER

There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Last Answer : NEED ANSWER

The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Description : The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Last Answer : Solution of this question

There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Last Answer : Find the mean of the given number is

Mean of 50 observations was found to be 80.4. -Maths 9th

Description : Mean of 50 observations was found to be 80.4. -Maths 9th

Last Answer : Here, n = 50, x̅ = 80.4 So, x̅ = ∑ xi/n ⇒ 80.4 = ∑ xi/50 ⇒ ∑ xi = 80.4 x 50 = 4020 Correct value of ∑ xi = 4020 - 69 + 96 = 4047 Correct mean = Correct value of ∑ xi/n = 4047/50 = 80.94

p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Description : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Last Answer : p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

What is 'SSS Congruence ' , 'C.P.C.T' ,'SAS' ?? -Maths 9th

Description : What is 'SSS Congruence ' , 'C.P.C.T' ,'SAS' ?? -Maths 9th

Last Answer : Side-Side-Side (SSS) Rule Side-Side-Side is a rule used to prove whether a given set of triangles are congruent. The SSS rule states that If three sides of one triangle are ... ASA, SSS and RHS congruence rules then remaining angles and sides corresponding in both triangles are equal.Example:

Express 1.27 bar in p/q form. -Maths 9th

Description : Express 1.27 bar in p/q form. -Maths 9th

Last Answer : (1) x = 1.2727272727......... (2) 100x = 127.27272727......... From subtracting (1) and (2 ) We get : 99x = 126.0000000 x = 126/99 = 14/11.

Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th

Description : Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th

Last Answer : p(x) = x4 - 3x2 + 2x - 5 According to remainder theorem, the required remainder will be = p(2) p(x) = x4 - 3x2 + 2x - 5 ∴ p(2) = 24 - 3(2)2 + 2(2) - 5 =16 - 12 + 4 - 5 = 3

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Description : ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Last Answer : (i) Since diagonals of a parallelogram bisect each other. ∴ O is the mid - point AC as well as BD. In △ADC, OD is a median. ∴ ar(△ADO) = ar(△CDO) [∵ A median of a triangle divide it into two triangles of equal ... and (i) , we have ar(△AOB) - ar(△AOP) = ar(△BOC) - ar(△COP) ⇒ ar(△ABP) = (△CBP)

In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY // WX, show that : -Maths 9th

Description : In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY // WX, show that : -Maths 9th

Last Answer : ar (ZPY)=ar( ZXY) they lie between the same base and between the same parallels Similarly, ar(WZY)=ar(ZPY) ar(ZWX)=ar(XWY)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Description : Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Last Answer : Draw AM ⟂ BD and CL ⟂ BD. Now, ar(△APB) × ar(△CPD) = {1/2 PB × AM} × {1/2 DP × CL} = {1/2 PB × CL} × {1/2 DP × AM} ar(△BPC) × ar(△APD) Hence, ar(△APB) × ar(△CPD) = ar(△APD) × ar(△BPC)

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

If P (event E) = 0.47, then find P(not E). -Maths 9th

Description : If P (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

SIMplify (2x+p-q)raise to power 2 -(2x-p+q)raise to power2 -Maths 9th

Description : SIMplify (2x+p-q)raise to power 2 -(2x-p+q)raise to power2 -Maths 9th

Last Answer : 2p-2q+2 is answer of the question

The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Description : The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Last Answer : Value of 1.999 is given below .

If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Description : If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Last Answer : (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6

Zero of the polynomial p(x)=2x+5 is -Maths 9th

Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

Last Answer : (b) Given, p(x) = 2x+5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ -5/2 Hence, zero of the polynomial p(x) is -5/2.

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

Find p(0), p( 1) and p(-2) for the following polynomials -Maths 9th

Description : Find p(0), p( 1) and p(-2) for the following polynomials -Maths 9th

Last Answer : (i) Given, polynomial is p(x) = 10x - 4x2 - 3 On putting x = 0,1 and - 2, respectively in Eq. (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 p(1) = 10 (1) - 4 (1 )2 -3 = 10-4-3= 10-7= 3 and p(-2 ... = (-2 + 2)(-2 -2) =0 (-4) = 0 Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0.

Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th

Description : Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th

Last Answer : Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 For zeroes of polynomial, put p(x) = 0 (x – 2)2 – (x+ 2)2 = 0 (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] ⇒ (2x)(-4) = 0

Check whether p(x) is a multiple of g(x) or not -Maths 9th

Description : Check whether p(x) is a multiple of g(x) or not -Maths 9th

Last Answer : p(x) is a multiple of g(x) or not

By remainder theorem, find the remainder when p(x) is divided by g(x) -Maths 9th

Description : By remainder theorem, find the remainder when p(x) is divided by g(x) -Maths 9th

Last Answer : Find the remainder when p(x) is divided by g(x)

Show that p-1 is a factor of p10 -1 and also of p11 -1. -Maths 9th

Description : Show that p-1 is a factor of p10 -1 and also of p11 -1. -Maths 9th

Last Answer : Let g (p) = p10 -1 and h(p) = p11 -1. On putting p=1 in Eq. (i), we get g(1)=110-1= 1-1=0 Hence, p-1 is a factor of g(p). Again, putting p = 1 in Eq. (ii), we get h (1) = (1)11 -1 = 1 -1 = 0 Hence, p -1 is a factor of h(p).

The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Last Answer : p(x) is divided by x+ 2 =

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

In following figure, coordinates of P are -Maths 9th

Description : In following figure, coordinates of P are -Maths 9th

Last Answer : (b) Here, given point P lies in II quadrant, so its abscissa will be negative and ordinate wilt be positive. Also, its perpendicular distance from X-axis is 4, so y-coordinate of P is 4 and its perpendicular distance from Y-axis is 2, so x-coordinate is -2. Hence, coordinates of P are (-2, 4).

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Description : Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Last Answer : (c) We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Description : The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Last Answer : (a) We know that, abscissa or the x-coordinate of a point is its perpendicular distance from the Y-axis. So, perpendicular distance of the point P(3, 4)from Y-axis = Abscissa = 3.