Are the ratios 1/2 and 8/16 a proportion?

Mathematics

2 answers:

vredina [299]3 years ago

5 0

I'm pretty sure they are but i would get a second opinion.

Vesna [10]3 years ago

3 0

Yes. 1*2*2 and 8*2=16, therefore they are proportional

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Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0

Xelga [282]

Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
$\left(h+c,\:k\right),\:\left(h-c,\:k\right)$

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}$
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

$9x^2-y^2-36x-4y+23=0 \ \textgreater \ \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}$
$9x^2-36x-4y-y^2=-23$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms}$
$9\left(x^2-4x\right)-\left(y^2+4y\right)=-23$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1$
$\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \ \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \ Refine$
$\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1$

Now rewrite in hyperbola standardform
$\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1$

$\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3$
$\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)$

Now we must compute c
$\sqrt{1^2+3^2} \ \textgreater \ \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \ 1^2 = 1 \ \textgreater \ \sqrt{1+3^2}$

$3^2 = 9 \ \textgreater \ \sqrt{1+9} \ \textgreater \ \sqrt{10}$

Therefore the hyperbola foci is at $\left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)$

For the vertices we have $\left(2+1,\:-2\right),\:\left(2-1,\:-2\right)$

Simply refine it
$\left(3,\:-2\right),\:\left(1,\:-2\right)$
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0

4 years ago

(Please Answer Quickly)

iren2701 [21]

For an airplane we need to create a function. The airplane flies in straight line so this means that we will have linear function. The airplane starts left of the rainbow. So let's take the following function:
y=x+8

"Create a table of at least four values for the function that includes two points of intersection between the airplane and the rainbow."
To find the intersection points we need to solve system of two equations:
 $y=- x^{2} +36 \\ y=x+36$
Solution are two points:
(-1,35) and (0,36)
Now we can create a table
x -1 0 1 2
f(x) 35 36 37 38


"What is the domain and range of the rainbow? Explain what the domain and range represent. Do all of the values make sense in this situation? Why or why not?"

The domain represents all possible values that x can have. The range represents all possible values that y can have.
In this example not all values make sense. Rainbow goes above ground and touches it.
Limitations exist if we have fraction, root, logarithm, trigonometric function... We do not have any of these so there are no limitations, so x can have any value.
Domain: -6≤x≤6 because x can not have smaller values than those at ground touch.
The formula for parabola is:
$y=a x^{2} +b$
Depending on sign of $x^{2}$ the maximum or minimum value of y is equal to b. All other smaller or greater values are possible.
Range: 0≤y≤36 because y can not have smaller values than those at ground touch.

"What are the x- and y-intercepts of the rainbow? Explain what each intercept represents."
x- and y- intercepts represent points at which function touches x- and y- axis.
To find y- intercept we insert x=0 into equation:
y=0+36
y=36
To find x- intercept we insert y=0
0=- $x^{2}$ +36
$x^{2} =36 \\ x_{1} =-6 \\ x_{2} =6$

"Is the linear function you created with your table positive or negative? Explain."

Function is positive if it lies above x-axis. Function is negative if it lies below x-axis.
Positive on interval: x ∈ <-6,∞>
Negative on interval: x ∈ <-∞,-6>
Not defined: x= -6

"What are the solutions or solution to the system of equations created?Explain what it or they represent."

We already found these solutions:
$y=- x^{2} +36 \\ y=x+36$
Solution are two points:
(-1,35) and (0,36)
These solutions represent points of intersection of linear function and parabola.

"Explain, using complete sentences, the steps for graphing the function."

Step 1) we need to find domain and range
Step 2) we need to find x- and y- intercepts
Step 3) if the function is exponential we find minima and maxima
Step 4) we find at least three values for the function