3 years ago

The yearbook editor wants each page of the Activities section of the yearbook to have the

Mathematics

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

Step-by-step explanation:

since gcd(26, 39) = 13

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1.find the sum. (3^3+5x^2+3x-7)+(8x-6x^2+6)<br> 2. Find the Difference: (8x-4x^2+3\)-(x^3+7x^2+3x-8)

vlada-n [284]

$(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26$

or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):

$(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1$

$(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10$