Math, asked by PopoKopo, 1 month ago

Simplify by rationalizing the denominators : 1 / 2+ root 5 + 1 / root 5 - root 6 + 1 / root 7 + root 6 + 1 / root 7 + root 8

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Answered by ayanzubair

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)In right-angled triangle ABC, we have

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)In right-angled triangle ABC, we haveAB2+BC2=AC2

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)In right-angled triangle ABC, we haveAB2+BC2=AC2⇒(x−3)2+(y−4)2+(x−1)2+(y+1)2=(3−1)2+(4+1)2

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)In right-angled triangle ABC, we haveAB2+BC2=AC2⇒(x−3)2+(y−4)2+(x−1)2+(y+1)2=(3−1)2+(4+1)2⇒x2+y2−4x−3y−1=0 ...(ii)

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)In right-angled triangle ABC, we haveAB2+BC2=AC2⇒(x−3)2+(y−4)2+(x−1)2+(y+1)2=(3−1)2+(4+1)2⇒x2+y2−4x−3y−1=0 ...(ii)Substituting the value of x from (i) into (ii), we get

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)In right-angled triangle ABC, we haveAB2+BC2=AC2⇒(x−3)2+(y−4)2+(x−1)2+(y+1)2=(3−1)2+(4+1)2⇒x2+y2−4x−3y−1=0 ...(ii)Substituting the value of x from (i) into (ii), we get(423−10y)2+y2−(23−10y)−3y−1=0

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.Then, AB=BC⇒AB2=BC2⇒(x−3)2+(y−4)2=(x−1)2+(y+1)2⇒4x+10y−23=0⇒x=423−10y ....(i)In right-angled triangle ABC, we haveAB2+BC2=AC2⇒(x−3)2+(y−4)2+(x−1)2+(y+1)2=(3−1)2+(4+1)2⇒x2+y2−4x−3y−1=0 ...(ii)Substituting the value of x from (i) into (ii), we get(423−10y)2+y2−(23−10y)−3y−1=0⇒4

Answered by BrainlyHearted

Answer:

Simplify by rationalizing the denominators : 1 / 2+ root 5 + 1 / root 5 - root 6 + 1 / root 7 + root 6 + 1 / root 7 + root 8Simplify by rationalizing the denominators : 1 / 2+ root 5 + 1 / root 5 - root 6 + 1 / root 7 + root 6 + 1 / root 7 + root 8Simplify by rationalizing the denominators : 1 / 2+ root 5 + 1 / root 5 - root 6 + 1 / root 7 + root 6 + 1 / root 7 + root 8Simplify by rationalizing the denominators : 1 / 2+ root 5 + 1 / root 5 - root 6 + 1 / root 7 + root 6 + 1 / root 7 + root 8

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