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rational function
Given the rational expression $f(x)=\dfrac{(2x+3)(x-1)}{x-1}\Rightarrow f(x)=2x+3$ [$\because$ cancel out $x-1$ from both the nimerator and the denominator.] $x$-intercept is the same as the zeros of the function, and the $x$-intercept is the value of $x$ when $f(x)=0$. Thus, $2x+3=0\RightarrRead more
Given the rational expression [ cancel out from both the nimerator and the denominator.]
-intercept is the same as the zeros of the function, and the -intercept is the value of when . Thus, .
The -intercept is the value of when . Thus, .
Therefoe, the -intercept and the zero is at and the -intercept is at .
See lessrational equation
Given $\dfrac{1}{4x+3}+\dfrac{2}{4x+3}=\dfrac{10}{2x-5}$. Because the left-hand side of the equation have the same denominator, we can add the numerator and have, $\dfrac{3}{4x+3}=\dfrac{10}{2x-5}$ Multiplying both sides by the LCD, (4x+3)(2x-5), we have $(4x+3)(2x-5)\dfrac{3}{4x+3}=(4x+3)(2x-5)\dfrRead more
Given .
Because the left-hand side of the equation have the same denominator, we can add the numerator and have,
Multiplying both sides by the LCD, (4x+3)(2x-5), we have
.
See lessSolving Parabola
The variable having squared determines the opening of a parabola. If $x$ is the squared variable, then the parabola opens up or down but if $y$ is the squared variable, then the parabola opens left or right. Here, the y variable is squared and the coefficient is negative, therefore, the parabola opeRead more
The variable having squared determines the opening of a parabola. If is the squared variable, then the parabola opens up or down but if is the squared variable, then the parabola opens left or right. Here, the y variable is squared and the coefficient is negative, therefore, the parabola opens to the left.
The vertex form of the equation of a left-opened parabola with vertex (h, k), is given by or , where is the distance from the vertex to the focus of the parabola. Given . Therefore, the vertex is at (3, -4).
The focus of a left-opening parabola with vertex (h, k), is given by . Thus the focus of the given parabola is at .
The directrix of a left-opening parabola with vertex (h, k), is given by . Thus the directrix of the given parabola is .
See lessSolving Parabola
The variable having squared determines the opening of a parabola. If $x$ is the squared variable, then the parabola opens up or down but if $y$ is the squared variable, then the parabola opens left or right. Here, the y variable is squared and the coefficient is positive, therefore, the parabola opeRead more
The variable having squared determines the opening of a parabola. If is the squared variable, then the parabola opens up or down but if is the squared variable, then the parabola opens left or right. Here, the y variable is squared and the coefficient is positive, therefore, the parabola opens to the right.
The vertex form of the equation of a right-opened parabola with vertex (h, k), is given by or , where is the distance from the vertex to the focus of the parabola. Given . Therefore, the vertex is at (-3, -2).
The focus of a right-opened parabola with vertex (h, k), is given by . Thus the focus of the given parabola is at .
The axis of symmetry of a right-opened parabola with vertex (h, k), is given by . Thus the axis of symmetry of the given parabola is at .
The length of the latus rectum of a parabola is given by . Thus the length of the latus rectum of the given parabola is at .
See lessSolving Ellipse
The standard form of the equation of an ellipse centered at (h, k) with the major axis at the x-axis is given by $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$, where $a>b$. Comparing $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ and $\dfrac{x^2}{64}+\dfrac{y^2}{55}=1$, we can see that h = 0 and kRead more
The standard form of the equation of an ellipse centered at (h, k) with the major axis at the x-axis is given by , where .
Comparing and , we can see that h = 0 and k = 0. Therefore, the center of the ellipse is at the origin, (0, 0).
The coordinates of the vertices of an ellipse centered at (0, 0) and with the major axis in the x-axis is given by . In this case, the vertices are at points and . That is, and .
The equations of the directrices of an ellipse centered at (0, 0) with the major axis in the x-axis is given by , where is the eccentricity of the ellipse given by , where is the length of the foci obtained by . Here, . Thus, . Therefore, the equations of the directrices are .
The length of the latus rectum, , of an ellipse centered at (0, 0) and with the major axis in the x-axis is given by .
See lessSolving Ellipse
The standard form of the equation of an ellipse centered at (h, k) with the major axis at the y-axis is given by $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$, where $a>b$. Comparing $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ and $\dfrac{x^2}{9}+\dfrac{y^2}{36}=1$, we can see that h = 0 and kRead more
The standard form of the equation of an ellipse centered at (h, k) with the major axis at the y-axis is given by , where .
Comparing and , we can see that h = 0 and k = 0. Therefore, the center of the ellipse is at the origin, (0, 0).
The coordinates of the vertices of an ellipse centered at (0, 0) and with the major axis in the y-axis is given by . In this case, the vertices are at points and . That is, and .
The equations of the directrices of an ellipse centered at (0, 0) with the major axis in the y-axis is given by , where is the eccentricity of the ellipse given by , where is the length of the foci obtained by . Here, . Thus, . Therefore, the equations of the directrices are .
The length of the latus rectum, , of an ellipse centered at (0, 0) and with the major axis in the y-axis is given by .
See lessSolving Ellipse
The standard form of the equation of an ellipse centered at (h, k) with the major axis at the y-axis is given by $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$, where $a>b$. Comparing $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ and $\dfrac{x^2}{9}+\dfrac{y^2}{55}=1$, we can see that h = 0 and kRead more
The standard form of the equation of an ellipse centered at (h, k) with the major axis at the y-axis is given by , where .
Comparing and , we can see that h = 0 and k = 0. Therefore, the center of the ellipse is at the origin, (0, 0).
The coordinates of the vertices of an ellipse centered at (0, 0) and with the major axis in the y-axis is given by . In this case, the vertices are at points and .
The equations of the directrices of an ellipse centered at (0, 0) with the major axis in the y-axis is given by , where is the eccentricity of the ellipse given by , where is the length of the foci obtained by . Here, . Thus, . Therefore, the equations of the directrices are .
Thelength of the latus rectum, , of an ellipse centered at (0, 0) and with the major axis in the y-axis is given by .
See lessWhat is the answer in this trigonometry problem?
Given $\cos45=\dfrac{10}{x}$ Using the 45-45-90 special right triangle, $\cos45=\dfrac{1}{\sqrt{2}}$. Thus, $\dfrac{1}{\sqrt{2}}=\dfrac{10}{x}$ Multiplying both sides by the least common denominator (LCD), $x\sqrt{2}$, we have $x\sqrt{2}\times\dfrac{1}{\sqrt{2}}=x\sqrt{2}\times\dfrac{10}{x}$ $\RightRead more
Given
Using the 45-45-90 special right triangle, .
Thus,
Multiplying both sides by the least common denominator (LCD), , we have
.
See lessMathematics: Trigonometric functions
a.) The angle $\theta=\dfrac{11\pi}{4}$ is greater than $2\pi$ so $\dfrac{11\pi}{4}\equiv\dfrac{11\pi}{4}-2\pi=\dfrac{3\pi}{4}$. $\dfrac{3\pi}{4}$ is in the second quadrant so sine and cosecant are positive while other trigonometric functions are negative. The reference angle for $\dfrac{3\pi}{4}$,Read more
a.) The angle is greater than so .
is in the second quadrant so sine and cosecant are positive while other trigonometric functions are negative.
The reference angle for , is .
Thus, , , .
Similarly, , , .
b.) is in the fourth quadrant so cosine and secant are positive while other trigonometric functions are negative.
The reference angle for , is .
Thus, , , .
Similarly, , , .
NB. Use the special right triangles (30-60-90 and 45-45-90 right triangles) to get the trigonometric ratios.
See lessWhat is 7/9z⁰?
A law of exponents states that zero exponent equals 1. Thus, given . Therefore, .
A law of exponents states that zero exponent equals 1.
Thus, given .
Therefore, .
See less