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x(1400 x+6)+2(2x+4)
The expression shown above can be expressed as cx^(2)+kx+m, where 
c,k, and m are constants. What is the value of k?
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x(1400x+6)+2(2x+4)x(1400 x+6)+2(2x+4)\newlineThe expression shown above can be expressed as cx2+kx+mcx^{2}+kx+m, where cc, kk, and mm are constants. What is the value of kk?\newline_____

Full solution

Q. x(1400x+6)+2(2x+4)x(1400 x+6)+2(2x+4)\newlineThe expression shown above can be expressed as cx2+kx+mcx^{2}+kx+m, where cc, kk, and mm are constants. What is the value of kk?\newline_____
  1. Distribute 22 across expression: Now we have 1400x2+6x1400x^2 + 6x.\newlineNext, distribute the 22 across (2x+4)(2x + 4).\newlineCalculation: 2×2x+2×42 \times 2x + 2 \times 4
  2. Combine distributed terms: That gives us 4x+84x + 8. Now, let's add the two parts together: 1400x2+6x+4x+81400x^2 + 6x + 4x + 8. Calculation: 1400x2+(6x+4x)+81400x^2 + (6x + 4x) + 8
  3. Combine like terms: Combine like terms to find the coefficient of xx.\newlineCalculation: 6x+4x=10x6x + 4x = 10x
  4. Final expression: So the expression becomes 1400x2+10x+81400x^2 + 10x + 8. The coefficient of xx, which is kk, is 1010.

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