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Topics
Pre-Algebra
Mean
Mode
Greatest Common Factor
Least Common Multiple
Order of Operations
Fractions
Mixed Fractions
Prime Factorization
Exponents
Radicals
Algebra
Combine Like Terms
Solve for a Variable
Factor
Expand
Evaluate Fractions
Linear Equations
Quadratic Equations
Inequalities
Systems of Equations
Matrices
Trigonometry
Simplify
Evaluate
Graphs
Solve Equations
Calculus
Derivatives
Integrals
Limits
Algebra Inputs
Trigonometry Inputs
Calculus Inputs
Matrix Inputs
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Evaluate
\frac{ba^{5}}{2}
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Solution Steps
\frac{a^6b^2}{2ab}
Cancel out ab in both numerator and denominator.
\frac{ba^{5}}{2}
Differentiate w.r.t. a
\frac{5ba^{4}}{2}
Quiz
Algebra
5 problems similar to:
\frac{a^6b^2}{2ab}
Similar Problems from Web Search
18a^3b^2/2ab
http://www.tiger-algebra.com/drill/18a~3b~2/2ab/
18a3b2/2ab Final result : 9a4b3 Step by step solution : Step 1 : b2 Simplify —— 2 Equation at the end of step 1 : b2 (((18 • (a3)) • ——) • a) • b 2 Step 2 :Equation at the end of step 2 : b2 ...
(18a^3b^2)/(2ab^2)
https://www.tiger-algebra.com/drill/(18a~3b~2)/(2ab~2)/
(18a3b2)/(2ab2) Final result : 9a2 Step by step solution : Step 1 :Equation at the end of step 1 : Step 2 :Equation at the end of step 2 : Step 3 : (2•32a3b2) Simplify —————————— 2ab2 ...
(3a^3b^2/2ab)^-2
https://www.tiger-algebra.com/drill/(3a~3b~2/2ab)~-2/
(3a3b2/2ab)(-2) Final result : a(-8)b(-6) • 22 ——————————————— 1 • 32 Reformatting the input : Changes made to your input should not affect the solution: (1): "^-2" was replaced by "^(-2)". Step by ...
is there any analytical way to konw if \frac{1}{2x}+\frac{x}{2} >1 for (1,\infty) or (0,\infty)?
https://math.stackexchange.com/questions/2388674/is-there-any-analytical-way-to-konw-if-frac12x-fracx2-1-for-1-in
Note that 0\leq (a-b)^2 = a^2 - 2ab + b^2 and hence a^2+b^2 \geq 2ab. Therefore, assuming ab>0, we have \frac{a^2+b^2}{2ab} \geq 1.
Reducing fractions?
https://math.stackexchange.com/q/60726
For the first fraction: \begin{align} \frac{2x + 2y}{x + y} &= \frac{2(x + y)}{x + y} \\ &= 2 \text{ assuming } (x+y) \neq 0 \text{ and dividing both numerator and denominator by (x + y)} \end{align} ...
Is there a pair of numbers a,b\in\Bbb{R} such that \frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}?
https://math.stackexchange.com/questions/2402803/is-there-a-pair-of-numbers-a-b-in-bbbr-such-that-frac1ab-frac1a
A simple proof for a^2 + ab + b^2 \neq 0 for non-zero reals a and b is as follows. 2(a^2+ab+b^2) = (a+b)^2 + a^2 + b^2=0 implies a=b=0. Hence, a contradiction.
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\frac{ba^{5}}{2}
Cancel out ab in both numerator and denominator.
Similar Problems
x \cdot x^2 \cdot 3x
n^4 \cdot 2n^2 \cdot n^5
(2a \cdot 3b^2)^2 \cdot c \cdot (2bc^3)^3
\frac{a^6b^2}{2ab}
\frac{x^3y^5}{3x} \times \frac{y^4}{x^2}
\frac{x^3y^5}{3x} \div \frac{y^4}{x^2}
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