602 results for "solve".
-
Question in Bill's workspace
Solve for $x$: $c(a^2)^x + d(a)^{x+1}=b$ (there is only one solution for this example).
-
Question in Bill's workspace
Solve for $x$: $a\cosh(x)+b\sinh(x)=c$. There are two solutions for this example.
-
Question in Bill's workspace
Solve for $x$: $\displaystyle ax ^ 2 + bx + c=0$.
-
Question in Bill's workspace
Differentiate $x^m\cos(ax+b)$
-
Question in Bill's workspace
Differentiate $ (a+bx) ^ {m} \sin(nx)$
-
Question in Bill's workspace
Differentiate $ (ax+b)^m(cx+d)^n$ using the product rule. The answer will be of the form $(ax+b)^{m-1}(cx+d)^{n-1}g(x)$ for a polynomial $g(x)$. Find $g(x)$.
-
Question in Bill's workspace
Differentiate $ x ^ m(ax+b)^n$ using the product rule. The answer will be of the form $x^{m-1}(ax+b)^{n-1}g(x)$ for a polynomial $g(x)$. Find $g(x)$.
-
Question in Bill's workspace
Differentiate $f(x) = x^m(a x+b)^n$.
-
Question in Bill's workspace
Differentiate $ x ^m \sqrt{a x+b}$.
The answer is in the form $\displaystyle \frac{x^{m-1}g(x)}{2\sqrt{ax+b}}$
for a polynomial $g(x)$. Find $g(x)$. -
Question in Bill's workspace
Differentiate the function $(a + b x)^m e ^ {n x}$ using the product rule.
-
Question in Bill's workspace
Inverse and division of complex numbers. Four parts.
-
Question in Bill's workspace
Multiplication of complex numbers. Four parts.
-
Question in Bill's workspace
Find $p$ and $q$ such that $ax^2+bx+c = a(x+p)^2+q$.
Hence, or otherwise, find roots of $ax^2+bx+c=0$.
Includes a video which shows how to solve a quadratic by completing the square.
-
Exam (4 questions) in mathcentre
4 questions on using partial fractions to solve indefinite integrals.
-
Question in Content created by Newcastle University
Find the solution of a constant coefficient second order ordinary differential equation of the form $ay''+by=0$. Complex roots.
-
Question in Kieran's workspace
Solve 4 first order differential equations of two types:$\displaystyle \frac{dy}{dx}=\frac{ax}{y},\;\;\frac{dy}{dx}=\frac{by}{x},\;y(2)=1$ for all 4.
rebelmaths
-
Question in PV EnglishFormulate a recurrence relation (= difference equation) and solve this recurrence relation.
-
Question in Clare Differential Equations
Solve: $\displaystyle \frac{d^2y}{dx^2}+2a\frac{dy}{dx}+a^2y=0,\;y(0)=c$ and $y(1)=d$. (Equal roots example).
-
Question in All questions
Some quadratics are to be solved by factorising
-
Question in All questions
A quadratic equation (equivalent to $(x+a)^2-b$) is given and sketched. Three equations are given that can be solved using the graph. There is a chance there will only be one solution.
-
Question in All questions
A few quadratic equations are given, to be solved by completing the square. The number of solutions is randomised.
-
Question in Ben's workspace
Solve the equation 5x - 8 = 32
-
Question in John's workspace
Some quadratics are to be solved by factorising
-
Question in Archive
Solve 4 first order differential equations of two types:$\displaystyle \frac{dy}{dx}=\frac{ax}{y},\;\;\frac{dy}{dx}=\frac{by}{x},\;y(2)=1$ for all 4.
rebelmaths
-
Question in Ruth's workspace
Using BIDMAS rules to solve equations
-
Question in All questions
Some quadratics are to be solved by factorising
-
Question in Content created by Newcastle University
Calculate a repeated integral of the form $\displaystyle I=\int_0^1\;dx\;\int_0^{x^{m-1}}mf(x^m+a)dy$
The $y$ integral is trivial, and the $x$ integral is of the form $g'(x)f'(g(x))$, so it straightforwardly integrates to $f(g(x))$.
-
Exam (1 question) in Content created by Newcastle University
Solve a system of linear equations using Gaussian elimination.
-
Question in Content created by Newcastle University
Solving a pair of congruences of the form \[\begin{align}x &\equiv b_1\;\textrm{mod} \;n_1\\x &\equiv b_2\;\textrm{mod}\;n_2 \end{align}\] where $n_1,\;n_2$ are coprime.
-
Question in Content created by Newcastle University
Solving two simultaneous congruences:
\[\begin{eqnarray*} c_1x\;&\equiv&\;b_1\;&\mod&\;n_1\\ c_2x\;&\equiv&\;b_2\;&\mod&\;n_2\\ \end{eqnarray*} \] where $\operatorname{gcd}(c_1,n_1)=1,\;\operatorname{gcd}(c_2,n_2)=1,\;\operatorname{gcd}(n_1,n_2)=1$