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3 years ago

A rectangle with length n and height m. A right triangle with hypotenuse n + 1, length n, and height m.

Mathematics

1 answer:

Oksana_A [137]3 years ago

8 0

9514 1404 393

Answer:

C. 12cm

Step-by-step explanation:

The equation for the perimeter of the rectangle is ...

P = 2(L+W)

34 = 2(n +m)

Solving for m, we get

m = 17 -n . . . . . . . divide by 2, subtract n

The Pythagorean theorem gives the relationship between the sides and the hypotenuse

m^2 +n^2 = (n+1)^2

(17 -n)^2 +n^2 = (n +1)^2 . . . . . . substitute for m

289 -34n +n^2 +n^2 = n^2 +2n +1 . . . . eliminate parentheses

n^2 -36n +288 = 0 . . . . . . . put in standard form

(n -12)(n -24) = 0 . . . . . . . . . factor

n = 12 . . . . . . . . . . n=24 is an extraneous solution here

The value of n is 12 cm.

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How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

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Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

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Answer:

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Step-by-step explanation:

Y-intercept: The y-intercept is equivalent to the point where x= 0. 'x' is the input variable in an equation, therefore the y-intercept is where the input, or x, is equal to 0.

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X-intercept: The x-intercept is the location of the graph where y= 0, or the output is equal to 0.

f(x) >0: In this, there is a 'greater than' sign. This means that f(x), or 'y', is greater than 0. Therefore, this consists of intervals of the domain above the x-axis.

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