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2 years ago

Use the sequence of operations to answer the question.

Mathematics

1 answer:

velikii [3]2 years ago

6 0

Answer:

(8-(3+b))3

Step-by-step explanation:

first of all, we have to variables, 3 and b that we are told to add

that is, 3+b

we are now told to subtract the answer from 8

that is 8 minis the answer

= 8- (3+b)

lastly we triple the answer to give

(8-(3+b)3

3(8-(3+b)

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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!

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3 years ago

Max ran 50 meters in 10 seconds. molly ran 30 meters in 5 seconds. who ran faster

Stolb23 [73]

To compare their speed (rate), we need to have the same time or same distance to use as a reference.

Matt ran 50 meters per 10 seconds. Molly ran 30 meters per 5 seconds. If she ran for 5 more seconds, she would have gone 30 more meters, for 30 meters per 10 seconds. Thus, Molly went faster.

Alternately, you could calculate their speeds in meters per (one) second and compare. 50 meters per 10 seconds is 5 meters per second, and 30 meters per 5 seconds is 6 meters per second.

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Distance from sun in au from mercury pls thanks

mestny [16]

<span>0.39 AU, 36 million miles 57.9 million km

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3 years ago

Calculate the number of acres required to produce 1,000 gallons of oil in one year from microalgae and soybeans. Which one (micr

Inessa05 [86]

Answer:

microalgae required less space.

Step-by-step explanation:

Amount of oil to be produced = 1,000 gallons

Now,

We know

10,000 gallons of oil is extracted from the miroalgae per acre per year

and,

From soyabeans the oil extracted = 50 gallons per acre per year

therefore,

To produce 1,000 gallons number of acres required by microalgae

= [ Amount of oil to be produced ] ÷ [ oil extracted per acre from microalgae]

= 1,000 ÷ 10,000

= 0.1 acres

similarly

To produce 1,000 gallons number of acres required by soyabeans

= [ Amount of oil to be produced ] ÷ [ oil extracted per acre from soyabeans]

= 1,000 ÷ 50

= 20 acres

hence,

microalgae required less space.

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Answer:

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