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What is the greater of two consecutive negative integers whose product is 132?
Let x = the lesser integer and let x + 1 = the greater integer. Because product is a key word for multiplication, the equation is x(x + 1) = 132. Multiply by using the distributive property on the left side of the equation: x2 + x = 132. Put the equation in standard form and set it equal to zero: x2 + x - 132 = 0. Factor the trinomial: (x - 11)(x + 12) = 0. Set each factor equal to zero and solve: x - 11 = 0 or x + 12 = 0; x = 11 or x = -12. Since you are seems for a negative integer and reject the x-value of 11. Thus, x = -12 and x + 1 = -11. The greater negative integer is -11.
The general solution of the differential equation (dy/dx) +x^2 = x^2*e^(3y). Solution)(dy/dx) +x^2 = x^2*e^(3y) dy/dx=x 2 (e 3y -1) x 2 dx=dy/(e 3y -1) this is an elementar
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use the distributive law to write each multiplication in a different way. then find the answer. 12x14 16x13 14x18 9x108 12x136 20x147
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