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2 years ago

What is the surface area of the figure

Mathematics

1 answer:

S_A_V [24]2 years ago

6 0

Answer:

you multiply the first the square Indian that will give you the essay equals in a square

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

son4ous [18]

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;

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

3 0

11 months ago

A is a 4x4 matrix and A^2 + 4A - 5I = 0. If det(A+2I)>0, enter det(A+2I)

maw [93]

$\mathbf A^2+4\mathbf A-5\mathbf I=0$
$\implies\mathbf A^2+4\mathbf A+4\mathbf I=(\mathbf A+2\mathbf I)^2=9\mathbf I$
$\implies\det\bigg((\mathbf A+2\mathbf I)^2\bigg)=\det(9\mathbf I)$
$\implies\det(\mathbf A+2\mathbf I)^2=9^4\det\mathbf I$
$\implies\det(\mathbf A+2\mathbf I)=\pm\sqrt{9^4}$
$\implies\det(\mathbf A+2\mathbf I)=9^2=81$

where we take the positive root because we're told that the determinant is positive.

5 0

2 years ago

What is the x-coordinate of the soultion y=x+2 y=-x+6

Kay [80]

<span>x+2 = -x+6
</span>2x+2 = 6
2x = 4
x = 2

6 0

3 years ago

Read 2 more answers

In the figures below, the cube-shaped box is 6 inches wide and the rectangular box is 10 inches long, 4 inches wide, and 4 inche

11111nata11111 [884]

The difference in volume= $v^{3} -V*L*H$
= $6^{3}-(10*4*4)$
=216-160
= 56 cubic inches

4 0

3 years ago

Suppose the operators 1,2,3,4 on ℝ^2 are given by

andrew11 [14]

Answer:

Two-Stage Least Squares (2SLS) Regression Analysis

Two-Stage least squares (2SLS) regression analysis is a statistical technique that is used in the analysis of structural equations. This technique is the extension of the OLS method. It is used when the dependent variable’s error terms are correlated with the independent variables. Additionally, it is useful when there are feedback loops in the model. In structural equations modeling, we use the maximum likelihood method to estimate the path coefficient. This technique is an alternative in SEM modeling to estimate the path coefficient. This technique can also be applied in quasi-experimental studies.

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Linear Transformation

6.1 Intro. to Linear Transformation

Homework: [Textbook, §6.1 Ex. 3, 5, 9, 23, 25, 33, 37, 39 ,53, 55,

57, 61(a,b), 63; page 371-].

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6 0

1 year ago