In order to get the result we have to find the HCF of 2002 and 2618. which is given by
2002 = 2 {tex}\times{/tex} 7 {tex}\times{/tex} 11 {tex}\times{/tex} 13 and
2618 = 2 {tex}\times{/tex} 7 {tex}\times{/tex} 11 {tex}\times{/tex} 17
Hence  HCF = 2 {tex}\times{/tex} 7 {tex}\times{/tex} 11 = 154

According to the question, given equation is {tex} 2x + 3y + 15 = 0.{/tex}
So, putting x = 2k - 3 and y = k + 2 in equation, we get
{tex}\Rightarrow{/tex} {tex}2(2k - 3) + 3(k + 2) +15 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}4k - 6+ 3k + 6 + 15 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}7k + 15 = 0{/tex}
{tex} \Rightarrow 7 k=-15 \\ \Rightarrow \ k=-\frac{15}{7}{/tex}

We need to prove that every line segment has one and only one mid-point. Let us consider the given below line segment AB and assume that C and D are the mid-points of the line segment AB

If C is the mid-point of line segment AB, then
AC = CB.
An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.”
AC + AC = CB + AC....(i)
From the figure, we can conclude that CB + AC will coincide with AB.
An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” AC + AC = AB....(ii)
An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.”
Let us compare equations (i) and (ii), to get
AC + AC = AB, or 2AC = AB.(iii)
If D is the mid-point of line segment AB, then
AD = DB.
An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.”
AD + AD = DB + AD....(iv)
From the figure, we can conclude that DB + AD will coincide with AB.
An axiom of the Euclid says that “Things which coincide with one another are equal to one another.”
AD + AD = AB....(v)
An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.”
Let us compare equations (iv) and (v), to get
AD + AD = AB, or
2AD = AB....(vi)
An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” Let us compare equations (iii) and (vi), to get
2AC = 2AD.
An axiom of the Euclid says that “Things which are halves of the same things are equal to one another.” AC = AD.
Therefore, we can conclude that the assumption that we made previously is false and a line segment has one and only one mid-point.

Given, AC = BC
AC + AC = BC + AC . . . . [AC are added to both the side]
2AC = AB . . . . [BC + AC coincides with AB]
{tex}\therefore{/tex} AC = {tex}\frac{1}{2}{/tex}AB

{tex}95 \times 96{/tex}
{tex}95 \times 96{\text{ can also be written as}}\left( {100 - 5} \right)\left( {100 - 4} \right){/tex} We can observe that we can apply the identity {tex}\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab{/tex}

Here a = -5 and b = -4
{tex}\left( {100 - 5} \right)\left( {100 - 4} \right) = {\left( {100} \right)^2} + \left[ {\left( { - 5} \right) + \left( { - 4} \right)} \right]\left( {100} \right) + \left( { - 5} \right) \times \left( { - 4} \right){/tex}
 {tex}= 10000 - 900 + 20{/tex}
= 9120 Therefore, we conclude that the value of the product {tex}95 \times 96{/tex} is 9120

You can check NCERT Solutions here : https://mycbseguide.com/ncert-solutions.html

Given: A ray OA.
Required: To construct an angle of 135⁰ at O.
Steps of construction :

1. Produce AO to A' to form OA'.
2. Taking O as centre and some radius, draw an arc of a circle, which intersects OA at a point B and OA' at a point B'.
3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.
4. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
   
5. Draw the ray OE passing through C.
   Then {tex}\angle{/tex}EOA = 60^o.
6. Draw the ray OF passing through D.
   Then {tex}\angle{/tex}FOE = 60^o.
7. Taking C and D as centres and with the radius more than {tex}1 \over2{/tex}CD, draw arcs to intersect each other, say at G.
8. Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE, i.e. {tex}\angle{/tex}FOG = {tex}\angle{/tex}EOG = {tex}1 \over2{/tex}{tex}\angle{/tex}FOE = {tex}1 \over2{/tex}(60^o) = 30^o.
   Thus {tex}\angle{/tex}GOA = {tex}\angle{/tex}GOE + {tex}\angle{/tex}EOA = 30^o + 60^o = 90^o
   {tex}\angle{/tex}B'OH = 90^o
9. Taking B' and H as centres and with the radius more than {tex}1 \over2{/tex}B'H, draw arcs to intersect each other, say at I.
10. Draw the ray OI. This ray OI is the bisector of the angle B'OG i.e. {tex}\angle{/tex}B'OI = {tex}\angle{/tex}GOI = {tex}1 \over2{/tex}{tex}\angle{/tex}B'OG = {tex}1 \over2{/tex}(90^o) = 45^o
    {tex}\angle{/tex}IOA = {tex}\angle{/tex}IOG + {tex}\angle{/tex}GOA
    = 45^o + 90^o = 135^o
    On measuring the {tex}\angle{/tex}IOA by protractor, find that {tex}\angle{/tex}IOA = 135^o.

(7+ 5 +x+ 8 + 17 + 12)/5 =10
49 + x = 50
X = 50 - 49 = 1
 

2x² + px - 14  = 0
Put x = 2
2(2)² + p(2) - 14 = 0
2 (4) + 2p - 14 = 0
8 + 2p - 14 = 0
2p - 6 = 0
2p = 6
p = 6/2
 p = 3

{tex}x^6−y^6=(x−y)(x+y)(x^4+y^4+x^2y^2){/tex}

Explanation:

{tex}x^6−y^6{/tex}

<mathjax>={tex}(x^2)^3−(y^2)^3{/tex}</mathjax> 

<mathjax>={tex}(x^2−y^2)((x^2)^2+(y^2)^2+(x^2)(y^2)){/tex}</mathjax> 

<mathjax>={tex}(x^2−y^2)(x^4+y^4+x^2y^2){/tex}</mathjax> 

<mathjax>={tex}(x−y)(x+y)(x^4+y^4+x^2y^2){/tex}</mathjax>