
Solve j-8/j+10=j-1/j+3 | Microsoft Math Solver
Multiply both sides of the equation by \left(j+3\right)\left(j+10\right), the least common multiple of j+10,j+3. j^{2}-5j-24=\left(j+10\right)\left(j-1\right) Use the distributive property to multiply j+3 by j-8 and combine like terms.
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Solve 0j-10j+6j | Microsoft Math Solver
j2+2j-24=0 Two solutions were found : j = 4 j = -6 Reformatting the input : Changes made to your input should not affect the solution: (1): "j2" was replaced by "j^2". Step by step ...
Solve j^2+3j-10 | Microsoft Math Solver
j=\frac{-3±\sqrt{3^{2}-4\left(-10\right)}}{2} All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
Solve 2+4j+5j+10j^2 | Microsoft Math Solver
Addition and multiplication of ideals satisfies the distributive property, so it follows straightforwardly that: (I+J)^2=(I+J)(I+J)=I(I+J)+J(I+J)=I^2+IJ+JI+J^2
Solve 7-j=-3 | Microsoft Math Solver
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Solve 20+j20+10+j10= | Microsoft Math Solver
Here is a slightly different way to calculate the number of compositions of 40 generated from 2,5 and 10. Since a selection of 2,5 or 10 can be represented as \begin{align*} ...