All categories

3 years ago

What is this?!?!!? it makes no senss

Mathematics

1 answer:

12345 [234]3 years ago

5 0

Answer:

if a varible is by itself with no key it would mean a 1 but this would most likely mean multiplication although there is no parenthesies my awnser would be 6!

Step-by-step explanation:

but that would be the same as the question.

You might be interested in

PLEASE HELP

m_a_m_a [10]

The midpoint T equals to 3 and 4

3 0

3 years ago

Read 2 more answers

5c+16.5=13.5+10c<br> solve for c

lana66690 [7]

Answer:

c = 3/5

Step-by-step explanation:

5c+16.5=13.5+10c

Subtract 5c from both sides:

13.5+5c=16.5

5c=3

c = 3/5

7 0

4 years ago

Read 2 more answers

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

Can someone help me with section b please and thank you

blondinia [14]

Y=3x
It costs $3 per movie (x)

8 0

3 years ago

Madelyn's car used 6 gallons of gas to drive 126 miles. At what rate does her car use

Reil [10]

Answer:

Step-by-step explanation:

You are given GALLONS and MILES

you want to find MILES / GALLON

so just put those numebrs in

MILES / GALLON = 126 miles / 6 gallons = 21 m/g (or 21 MPG)

4 0

3 years ago