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3 years ago

Can you help me with this!!!!!!!!!

Mathematics

1 answer:

alexira [117]3 years ago

4 0

a. The first variable is x and the second variable is y.

b. The equations are $x+y=12$ and $x-y=8.$

Step-by-step explanation:

Step 1:

The first step is to define the variables. The variables can be any two symbols, letters, characters, etc.

Here let the first variable be x and the let the second variable be y.

So the variables are defined as x and y.

Step 2:

The sum of the given variables is 12.

The first variable + the second variable = 12,

$x+y=12.$

The difference between the two variables is 4.

The first variable - the second variable = 4.

$x-y=4.$

Step 3:

If we add both the equations we get, $2x =16, x= 8$ and $y=4.$

x = 8 and y = 4.

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A plot of land doubles in size by adding x meters to the length and x meters to the width of the land. if the original plot had

ladessa [460]

Lets write as an equation the info provided in the problem:
original area = 200*300 = 60000 squared meters
Adding x meters to the length is: 300 + x
Adding x meters to the width is: 200 + x
If the area doubles in size we have: 2*<span>60000
Now writing as a single equation all info:
(300 + x)(200 + x) = 120000
We have to make the operations and solve:
60000 + 300x + 200x + x^2 = </span><span>120000
x^2 + 500x - 60000 = 0
This is a squared trinomial, to solve it we need two numbers that subtracted give us 500 and multiplied -60000:
(x + 600)(x -100) = 0
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4 0

3 years ago

Please help please please

alexgriva [62]

Two sets (or three technically)
sets {2, 4, 6, 8, 10} & {8,9,10}

The probability of one of the above numbers because it is a union of those two vars/sets so numbers from either set go

{2, 4, 6, 8, 9, 10}
Thats 6 of the 10 numbers

6/10
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If i'm wrong, sorry, haven't done this kind of stuff in a while

8 0

3 years ago

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;

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

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2 years ago