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shusha [124]
4 years ago
9

Multiply (x2-5x)(2x2+x-3)​

Mathematics
2 answers:
Ulleksa [173]4 years ago
4 0

Answer:

2x^4-9x^3-8x^2+15x

Step-by-step explanation:

(x^2-5x)(2x^2+x-3)

2x^4+x^3-3x^2-10x^3-5x^2+15x

2x^4+x^3-10x^3-3x^2-5x^2+15x

2x^4-9x^3-8x^2+15x

Vesna [10]4 years ago
3 0

x^3 - 9x^2 + 23x - 12

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In a triangle ABC, AD is drawn perpendicular to BC. <br> Prove that AB2 - BD2 = AC2 - CD2.
Finger [1]

Answer:

Given: In triangle ABC , AD is drawn perpendicular to BC.

Since AD is drawn perpendicular to BC, it creates two right triangles: ADB and ADC.

Prove that: AB^2-BD^2 = AC^2-CD^2

Pythagoras triangle for right angle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

In a right angle triangle ADB;

AD^2+BD^2 = AB^2                 [By Pythagoras theorem]

or

AD^2= AB^2-BD^2                      .......[1]

Now, in right angle triangle ADC;

AD^2+CD^2 = AC^2                [By Pythagoras theorem]

or we can write this as;

AD^2= AC^2-CD^2                    ......[2]

Substituting the equation [1] in [2] we get;

AB^2-BD^2 =AC^2-CD^2            hence proved!

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