Help me find a laptop under 800$ (Incl. of taxes) that meets my requirements?

Help me find a laptop under 800$ (Incl. of taxes) that meets my requirements?
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Wal*Mart sells a Toshiba with those specs and at that price level.
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Description : When a raised face flange gasket surface is machined to a flat face surface, does the flange still comply with ASME B31.3? a) No. b) Yes. c) Yes, provided the flange meets thickness requirements for flat faced flanges per ASME B31.3. d) Yes, if approached by the flange committee.

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Description : Small and large size of greeting post rate is a) 14/- per card incl of INR 5/- postage, 17/- per card inclusive of INR 5/- postage. a) 12/- per card incl of INR 5/- postage, 16/- per card inclusive ... 5/- postage a) 10/- per card incl of INR 5/- postage, 12/- per card inclusive of INR 5/- postage

Last Answer : a) 14/- per card incl of INR 5/- postage, 17/- per card inclusive of INR 5/- postage.

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Description : Marketing effo ing efforts, wh rts, which incl ich include publi ude public health ca c health campaigns to reduc ns to reduce smoki e smoking, alcoh ng, alcoholism, drug abuse, and overeating are called _____. a. specialty product b. social marketing social marketing c. shopping product

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A man is walking down a road with a basket of eggs. As he is walking he meets someone who buys one-half of his eggs plus one-half of an egg. He walks a little further and meets another person who buys one-half of his eggs plus one-half of an egg. After proceeding further he meets another person who buys one-half of his eggs plus one half an egg. At this point he has sold all of his eggs, and he never broke an egg. How many eggs did the man have to start with? -Riddles

Description : A man is walking down a road with a basket of eggs. As he is walking he meets someone who buys one-half of his eggs plus one-half of an egg. He walks a little further and meets another person who buys ... of his eggs, and he never broke an egg. How many eggs did the man have to start with? -Riddles

Last Answer : 7 eggs. The first person bought one half of his eggs plus one half an egg (3 1/2 + 1/2 = 4 eggs) This left him 3 eggs. The second person bought one-half of his eggs plus one half an egg, (1 1/2 + ... last person bought one-half of his eggs plus one-half an egg, (1/2 + 1/2 = 1 egg) leaving no eggs.

A man is walking down a road with a basket of eggs. As he is walking he meets someone who buys one-half of his eggs plus one-half of an egg. He walks a little further and meets another person who buys one-half of his eggs plus one-half of an egg. After proceeding further he meets another person who buys one-half of his eggs plus one half an egg. At this point he has sold all of his eggs, and he never broke an egg. How many eggs did the man have to start with? -Riddles

Description : A man is walking down a road with a basket of eggs. As he is walking he meets someone who buys one-half of his eggs plus one-half of an egg. He walks a little further and meets another person who buys ... of his eggs, and he never broke an egg. How many eggs did the man have to start with? -Riddles

Last Answer : 7 eggs. The first person bought one half of his eggs plus one half an egg (3 1/2 + 1/2 = 4 eggs) This left him 3 eggs. The second person bought one-half of his eggs plus one half an egg, (1 1/2 + ... last person bought one-half of his eggs plus one-half an egg, (1/2 + 1/2 = 1 egg) leaving no eggs.

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

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)

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

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

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)

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Last Answer : NEED ANSWER

In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

Last Answer : This answer was deleted by our moderators...

Find the coordinate where the linear equation 3x -4y = 11 meets at x-axis. -Maths 9th

Description : Find the coordinate where the linear equation 3x -4y = 11 meets at x-axis. -Maths 9th

Last Answer : hope it helps

Find the coordinate where the linear equation 4x - 23 y = 7 meets at y-axis. -Maths 9th

Description : Find the coordinate where the linear equation 4x - 23 y = 7 meets at y-axis. -Maths 9th

Last Answer : 4x-2=-7*3y 4x+21y=2 The equation meets y axis when x=0 4.0+21y=2 y=21/2 Hence , the equation meets y-axis at (0,21/2)

In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th

Description : In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th

Last Answer : hope its clearhope its clear

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Description : The normal at a point `P` on the ellipse `x^2+4y^2=16` meets the x-axis at `Qdot` If `M` is the midpoint of the line segment `P Q ,` then the locus of

Last Answer : The normal at a point `P` on the ellipse `x^2+4y^2=16` meets the x-axis at `Qdot` If `M` is ... (pm2sqrt3,pmsqrt1/7)` D. `(pm2sqrt3,pm(4sqrt(3))/4)`

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Description : Find the equation of normal to the curve `y(x-2)(x-3)-x+7=0` at that point at which the curve meets X-axis.

Last Answer : Find the equation of normal to the curve `y(x-2)(x-3)-x+7=0` at that point at which the curve meets X-axis.

Where the water table meets the ground surface groundwater bubbles or flows out of cracks in the rock in openings called?

Description : Where the water table meets the ground surface groundwater bubbles or flows out of cracks in the rock in openings called?

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What is the distance from a point on the x axis to the centre of a circle when a tangent line at the point 3 4 meets the circle of x2 plus y2 -2x -6y plus 5 equals 0?

Description : What is the distance from a point on the x axis to the centre of a circle when a tangent line at the point 3 4 meets the circle of x2 plus y2 -2x -6y plus 5 equals 0?

Last Answer : Circle equation: x^2 +y^2 -2x -6y +5 = 0Completing the squares: (x-1)^2 +(y-3)^2 = 5Centre of circle: (1, 3)Tangent line meets the x-axis at: (0, 5)Distance from (0, 5) to (1, 3) = 5 units using the distanceformula

What is the equation of the tangent line that meets the circle x2 -4x plus y2 -6y equals 4 at the point 6 4?

Description : What is the equation of the tangent line that meets the circle x2 -4x plus y2 -6y equals 4 at the point 6 4?

Last Answer : Circle equation: x^2 -4x +y^2 -6y = 4Completing the squares: (x-2)^2 +(y-3)^2 = 17Point of contact: (6, 4)Center of circle: (2, 3)Slope of radius: 1/4Slope of tangent line: -4Tangent equation: y-4 = -4(x-6) => y = -4x+28Tangent line equation in its general form: 4x+y-28 = 0

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What is the point of contact when the line y equals 2x meets the circle x2 plus y2 -8x -y plus 5 equals 0?

Description : What is the point of contact when the line y equals 2x meets the circle x2 plus y2 -8x -y plus 5 equals 0?

Last Answer : If: y = 2xThen: y^2 = 4x^2If: x^2 +y^2 -8x -y +5 = 0Then: x^2 +4x^2 -8x -2x +5 = 0Transposing terms: 5x^2 -10x +5 = 0Dividing all terms by 5: x^2 -2x +1 = 0Factorizing the above: (x-1)(x-1) = 0 meaning x = 1By substitution into original equations point of contact is madeat: (1, 2)

What is the length of the tangent line from the point 8 2 to a point where it meets the circle x2 plus y2 -4x -8y -5 equals 0?

Description : What is the length of the tangent line from the point 8 2 to a point where it meets the circle x2 plus y2 -4x -8y -5 equals 0?

Last Answer : Equation of circle: x^2 +y^2 -4x -8y -5 = 0Completing the squares: (x-2)^2 +(y-4)^2 = 25 which is radiussquaredCenter of circle: (2, 4)Tangent line originates from: (8, ... angle triangleUsing Pythagoras theorem: distance^2 minus radius^2 = 15Therefore length of tangent line is the square root of 15

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What is the most critical aspect in developing a project plan that meets project specifications within the timeframe and at the lowest costs? Select one: a. Assessing risk managemenc. Emphasizing error prevention and customer satisfaction d. Requiring commitment to quality by all the employees and business partners e. Developing a project execution plan that matches the complexity level of the projectt b. Providing documents that specifically comply with the quality standards in use

Description : What is the most critical aspect in developing a project plan that meets project  specifications within the timeframe and at the lowest costs? Select one: a. Assessing risk management b ... business partners e. Developing a project execution plan that matches the complexity level of the project

Last Answer : e. Developing a project execution plan that matches the complexity level of the project

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The Inspector should, whenever he meets a village postman out in his beat or at the office to which he is attached, examine a) His bag to see whether he carries postage stamps, forms, etc., b) His book of receipts are pasted to the counterfoils of receipts for articles registered by him. c) Both the A and B d) Either A or B

Description : The Inspector should, whenever he meets a village postman out in his beat or at the office to which he is attached, examine a) His bag to see whether he carries postage stamps, forms, etc., b) His ... counterfoils of receipts for articles registered by him. c) Both the A and B d) Either A or B

Last Answer : c) Both the A and B