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3 years ago

The red arrow is pointing to what part of the xy-plane?

Mathematics

1 answer:

Brut [27]3 years ago

6 0

The answer is y axis

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Jacob is reading a great thriller. The table shows a proportional relationship between the number of pages he reads and the numb

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14 there are 7 days in one week

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What is the perimeter of a triangle with the side lengths 4 cm, 6 cm, and (3y-2) cm?

jek_recluse [69]

Answer:

Step-by-step explanation:

P = 4 + 6 + (3y - 2) Remove the brackets

P = 10 + 3y - 2 Combine

P = 8 + 3y

Note 3y must be less than 8. That's because 2 sides of any triangle must be larger than the 3rd side or else you don't have a triangle.

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Lynh manages a small company with 12 employees. He offers profit sharing where 5% of profits at end of year is distributed evenl

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3 years ago

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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!

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4 years ago

Which of the following are vertical asymptotes of the function y = 2cot(3x) + 4? Check all that apply. A.x = pi/3 B.x = +/- pi/2

Kisachek [45]

Vertical asymptotes occur when the denominator of a rational is 0, whilst not zeroing out the numerator, making the rational, undefined, in this case

$\bf y=2cot(3x)+4\implies y=2\cdot \cfrac{cos(3x)}{sin(3x)}+4\impliedby \textit{if \underline{sin(3x)} turns to 0}\\\\
-------------------------------\\\\
\textit{let's check}
\\\\\\
A)\qquad \cfrac{cos(3x)}{sin\left( 3\frac{\pi }{3} \right)}\implies \cfrac{cos(3x)}{sin\left( \pi \right)}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined$

$\bf B)\qquad \cfrac{cos(3x)}{sin\left( 3\frac{\pm\pi }{2} \right)}\implies\cfrac{cos(3x)}{sin\left( \frac{\pm3\pi }{2} \right)}\implies \cfrac{cos(3x)}{\pm 1}\implies \pm cos(3x)
\\\\\\
C)\qquad \cfrac{cos(3x)}{sin\left( 3(2\pi )\right)}\implies \cfrac{cos(3x)}{sin(6\pi )}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined
\\\\\\
D)\qquad \cfrac{cos(3x)}{sin(3(0))}\implies \cfrac{cos(3x)}{sin(0)}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined$

6 0

3 years ago

Read 2 more answers