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3 years ago

What is the correct answer

Mathematics

2 answers:

Ksju [112]3 years ago

4 0

The correct answer is C. If you reduce the fraction 36/18, it reduces to 2/1 or simply, 2. So, both fractions equal 2.

aksik [14]3 years ago

4 0

Answer: C

Step-by-step explanation:

Reduce the fraction 36/18 it is reduced to 2/1 so that means the answer would be 2

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For a special sale, a mens tie buyer plans to promote a $39.99 tie. Consequently, the buyer purchases 450 ties and wants to achi

zlopas [31]

Answer:

The average cost of each tie is $56.350

Step-by-step explanation:

We are given that a men tie buyer plans to promote a $39.99 tie. Consequently, the buyer purchases 450 ties

So, Cost of 450 ties = $450 \times 39.99=17995.5$

An order for 100 ties that cost $20.00 each

Cost of 100 ties = $100 \times 20 = 2000$

Total cost price = 17995.5+2000 = 19995.5

Let the selling price of 1 tie be x

Total ties = 450+100 = 550

So, SP of 550 ties =550x

Now we are given that markup is 55%

So, Profit % = 55%

So, $CP = \frac{SP \times 100}{100+P\%}$

Substitute the values

$19995.5 = \frac{550x \times 100}{100+55}$

$19995.5 = \frac{550x \times 100}{155}$

$19995.5 \times 155 = 55000x$

$\frac{19995.5 \times 155}{55000}=x$

$56.350=x$

Hence The average cost of each tie is $56.350

7 0

3 years ago

The graph of which function will have a maximum and a y-intercept of 4?

stiv31 [10]

Answer:

Step-by-step explanation:

We have quadratic functions f(x) = ax² + bx + c.

c - y-intercept

If a > 0, then a parabola opens up and has a minimum in a vertex.

If a < 0, then a parabola opens down and has a maximum in a vertex.

The function has maximum and y-intercept of 4:

a < 0 and c = 4

4 0

3 years ago

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!