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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
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Solve for h
\left\{\begin{matrix}h=\frac{x}{y}\text{, }&x\neq 0\text{ and }y\neq 0\\h\neq 0\text{, }&y=0\text{ and }x=0\end{matrix}\right.
{
h
=
y
x
,
h
=
0
,
x
=
0
and
y
=
0
y
=
0
and
x
=
0
View solution steps
Steps for Solving Linear Equation
y = h ^ { - 1 } ( x )
y
=
h
−
1
(
x
)
Swap sides so that all variable terms are on the left hand side.
Swap sides so that all variable terms are on the left hand side.
h^{-1}x=y
h
−
1
x
=
y
Reorder the terms.
Reorder the terms.
\frac{1}{h}x=y
h
1
x
=
y
Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by h.
Variable
h
cannot be equal to
0
since division by zero is not defined. Multiply both sides of the equation by
h
.
1x=yh
1
x
=
y
h
Swap sides so that all variable terms are on the left hand side.
Swap sides so that all variable terms are on the left hand side.
yh=1x
y
h
=
1
x
Reorder the terms.
Reorder the terms.
hy=x
h
y
=
x
The equation is in standard form.
The equation is in standard form.
yh=x
y
h
=
x
Divide both sides by y.
Divide both sides by
y
.
\frac{yh}{y}=\frac{x}{y}
y
y
h
=
y
x
Dividing by y undoes the multiplication by y.
Dividing by
y
undoes the multiplication by
y
.
h=\frac{x}{y}
h
=
y
x
Variable h cannot be equal to 0.
Variable
h
cannot be equal to
0
.
h=\frac{x}{y}\text{, }h\neq 0
h
=
y
x
,
h
=
0
Solve for x
x=hy,h\neq 0
x
=
h
y
,
h
=
0
View solution steps
Solution Steps
y = h ^ { - 1 } ( x )
y
=
h
−
1
(
x
)
Swap sides so that all variable terms are on the left hand side.
Swap sides so that all variable terms are on the left hand side.
h^{-1}x=y
h
−
1
x
=
y
Reorder the terms.
Reorder the terms.
\frac{1}{h}x=y
h
1
x
=
y
Multiply both sides of the equation by h.
Multiply both sides of the equation by
h
.
1x=yh
1
x
=
y
h
Reorder the terms.
Reorder the terms.
x=hy
x
=
h
y
Graph
Quiz
Linear Equation
5 problems similar to:
y = h ^ { - 1 } ( x )
y
=
h
−
1
(
x
)
Similar Problems from Web Search
How can I find T^{-1}(x,y,z) (inverted matrix) of a linear operator T:V_3 \to V_3
How can I find
T
−
1
(
x
,
y
,
z
)
(inverted matrix) of a linear operator
T
:
V
3
→
V
3
https://math.stackexchange.com/q/833500
Hint: Find the matrix of T wrt standard ordered basis,then find its inverse Solution: let a=(1,0,0),b=(1,1,0),c=(1,1,1) Hence T(a)=2a+b-c=(2,0,-1),T(b)=-b+2c=(1,1,2),T(c)=a+b-c=(1,0,-1) [This is ...
Hint: Find the matrix of
T
wrt standard ordered basis,then find its inverse Solution: let
a
=
(
1
,
0
,
0
)
,
b
=
(
1
,
1
,
0
)
,
c
=
(
1
,
1
,
1
)
Hence
T
(
a
)
=
2
a
+
b
−
c
=
(
2
,
0
,
−
1
)
,
T
(
b
)
=
−
b
+
2
c
=
(
1
,
1
,
2
)
,
T
(
c
)
=
a
+
b
−
c
=
(
1
,
0
,
−
1
)
[This is ...
Q_{30} Graph the curves y = x^3-4x and x=y^3-4y and find their points of intersection correct to one decimal place.
Q
3
0
Graph the curves
y
=
x
3
−
4
x
and
x
=
y
3
−
4
y
and find their points of intersection correct to one decimal place.
https://math.stackexchange.com/q/1910884
You just have to solve system y=x^3-4x x = y^3-4y by substitution. You get a 9th degree polynomial, and thus there are at most 9 intersections. By substituting we get x = (x^3-4x)^3-4(x^3-4x) \implies \underbrace{x^9-12 x^7+48 x^5-68 x^3+15 x}_{=: p(x)} = 0 ...
You just have to solve system
y
=
x
3
−
4
x
x
=
y
3
−
4
y
by substitution. You get a 9th degree polynomial, and thus there are at most 9 intersections. By substituting we get
x
=
(
x
3
−
4
x
)
3
−
4
(
x
3
−
4
x
)
⟹
=
:
p
(
x
)
x
9
−
1
2
x
7
+
4
8
x
5
−
6
8
x
3
+
1
5
x
=
0
...
Solving a general integral (expectation of some variant of exponential distribution)
Solving a general integral (expectation of some variant of exponential distribution)
https://math.stackexchange.com/questions/800086/solving-a-general-integral-expectation-of-some-variant-of-exponential-distribut
Actually the PDF of Y is the function g defined by g(y)=\lambda h(y)\mathrm e^{-\lambda H(y)}, where h is the derivative of H, if H is differentiable. In the general case, the CDF of Y ...
Actually the PDF of
Y
is the function
g
defined by
g
(
y
)
=
λ
h
(
y
)
e
−
λ
H
(
y
)
,
where
h
is the derivative of
H
, if
H
is differentiable. In the general case, the CDF of
Y
...
Minimization and Least Squares
Minimization and Least Squares
https://math.stackexchange.com/questions/213428/minimization-and-least-squares
Given line ax+by=0 and point (x_0,y_0), we want to find the distance from the point to the line. This distance is a perpendicular from the point (x_0,y_0) to given line. Rewriting line equation, ...
Given line
a
x
+
b
y
=
0
and point
(
x
0
,
y
0
)
, we want to find the distance from the point to the line. This distance is a perpendicular from the point
(
x
0
,
y
0
)
to given line. Rewriting line equation, ...
Random number distribution from a different distribution
Random number distribution from a different distribution
https://math.stackexchange.com/questions/906061/random-number-distribution-from-a-different-distribution
Long form of angryavian's argument: Let Y=\frac1aX=g(X), with X having pdf f_X, and a>0. There is a standard way to find the pdf of Y: for any integrable function h, E(h(Y))=\int_{-\infty}^\infty h(y) f_Y(y) \mathrm{d}y ...
Long form of angryavian's argument: Let
Y
=
a
1
X
=
g
(
X
)
, with
X
having pdf
f
X
, and
a
>
0
. There is a standard way to find the pdf of Y: for any integrable function
h
,
E
(
h
(
Y
)
)
=
∫
−
∞
∞
h
(
y
)
f
Y
(
y
)
d
y
...
Find area bounded by inverse of f(x)=x^3+3x+1,the ordinates at x=-3,x=5 and x-axis
Find area bounded by inverse of
f
(
x
)
=
x
3
+
3
x
+
1
,the ordinates at
x
=
−
3
,
x
=
5
and x-axis
https://math.stackexchange.com/q/2830453
First note that function is one-one and onto in the range (-3,5). So the inverse does exist. So if you want to find the area enclosed by f^{-1} which is the reflection of f(x) about the line y=x ...
First note that function is one-one and onto in the range
(
−
3
,
5
)
. So the inverse does exist. So if you want to find the area enclosed by
f
−
1
which is the reflection of
f
(
x
)
about the line
y
=
x
...
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h^{-1}x=y
Swap sides so that all variable terms are on the left hand side.
\frac{1}{h}x=y
Reorder the terms.
1x=yh
Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by h.
yh=1x
Swap sides so that all variable terms are on the left hand side.
hy=x
Reorder the terms.
yh=x
The equation is in standard form.
\frac{yh}{y}=\frac{x}{y}
Divide both sides by y.
h=\frac{x}{y}
Dividing by y undoes the multiplication by y.
h=\frac{x}{y}\text{, }h\neq 0
Variable h cannot be equal to 0.
h^{-1}x=y
Swap sides so that all variable terms are on the left hand side.
\frac{1}{h}x=y
Reorder the terms.
1x=yh
Multiply both sides of the equation by h.
x=hy
Reorder the terms.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
x
2
−
4
x
−
5
=
0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
4
sin
θ
cos
θ
=
2
sin
θ
Linear equation
y = 3x + 4
y
=
3
x
+
4
Arithmetic
699 * 533
6
9
9
∗
5
3
3
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
[
2
5
3
4
]
[
2
−
1
0
1
3
5
]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
{
8
x
+
2
y
=
4
6
7
x
+
3
y
=
4
7
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
d
x
d
(
x
−
5
)
(
3
x
2
−
2
)
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
∫
0
1
x
e
−
x
2
d
x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
x
→
−
3
lim
x
2
+
2
x
−
3
x
2
−
9
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