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shepuryov [24]
2 years ago
5

The table shows the relationship between the number of drops of food color added to different number of cups of cake frosting:

Mathematics
2 answers:
stepladder [879]2 years ago
6 0

I'd say that they are all equivalent because when you divided all of them it equals each other.

2:6 = 1/3

3:9 = 1/3

4:12 = 1/3

vitfil [10]2 years ago
6 0
I would say that it would be 1:3 since 4/12= 3 and 9/3=3 and 6/2=3
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What is the 11th term in the sequence with the nth term :
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Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!
IgorC [24]
Hey there, hope I can help!

-4x^2+9y^2+32x+36y-64=0

\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \  9y^2+32x+36y-4x^2=64

\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \  -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4
-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \  -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1

Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1

For me I used
\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1
As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1

Hope this helps!
4 0
3 years ago
Round 392,153 to the nearest hundred thousand
Galina-37 [17]
392,000 would be the nearest <span>hundred thousand</span>
6 0
3 years ago
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