Q: When 3a2-2a plus 5 subtracted by a2 plus a equals 1?

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5a+3a2a2= 8a a2

a2

l a2 b2 is c2!!Its completely norma

a=3

144 Formula: c= (a2)+(ab)

a right triangle

Pythagorean Theorem

a2+30a+56=0 , solve for a Using the quadratic formula, you will find that: a=-2 , a=-28

You just typed it.

a2+b2+c2=x2+y2+z2 divide each side by 2 (a2+b2+c2)/2=(x2+y2+z2)/2 a+b+c=x+y+z

It is an expression and a term that are of equal value

If I've read your question correctly, you need to subtract: a2 +2a -7 a2 -4a2 +5a2 -6 = 2a2 -6 Note, if x - y = z, then y = x - z; so: 2a2 -6 - (a2 -2a +1) = 2a2 -6 - a2 +2a -1 = a2 +2a -7

Pythagoras' theorem for a right angle triangle.

A2 + B2 = C2 If C=8, then A2 + B2 = 64

( a2 ) ( a2+1 )

This is known as the Cosine Rule.

a2 + -5a2 - a2 - a = a2 - 5a2 - a2 - a = a2 - 6a2 - a = -5a2 - a

a2 - 4a + 4

Pythagoras. That's why it's called the Pythagorean Theorem.

a2 + 62 = 122 a2 + 62 - 62 = 122 - 62 a2 = 144 - 36 a2 = 108 taking the square root of each side, we get a equal plus or minus the square root of 108, or plus or minus 6 times the square root of 3.

5 - a - 1 + c - 6a2 + a2

Rprime= 2q + 2a + 3 Rdoubleprime= 4

a2 + a2 = 2a2

a2+9a+20 = (a+4)(a+5) when factored

a2+2a2b+2ab2+b2

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