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2 years ago

For the dinner special at a restaurant, the customer must choose an appetizer, an entree, a side dish, and a dessert.

Mathematics

1 answer:

gtnhenbr [62]2 years ago

8 0

Answer:

3 dinner specials are possible to make.

Step-by-step explanation:

you group 4 things up, 1 from each category then you have your groups when you can't group 4!

Which results in the answer- 3 dinner specials!

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2n-3/5=5 solve for n

dezoksy [38]

N=14/5 thats the answer.

3 0

3 years ago

Read 2 more answers

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

The last time Sierra played softball, she hit the ball 45% of the times she was at bat. Based on this information, how many time

Ymorist [56]

Answer:

Number of times she will hit the ball the next time she plays softball if she is at bat 20 times = 9 times

Step-by-step explanation:

Percentage of times Sierra hit the ball when she was at bat playing softball= 45%

So, we can find:

Number of times she will hit the ball the next time she plays softball if she is at bat 20 times = $\textrm{45\% of 20 times} =0.45\times 20=9\ times$

3 0

3 years ago

What’s 1/3 • 5/4 and 1/5•2/3

Debora [2.8K]

Answer:

Step-by-step explanation:

1/2 + 2/3 + 5/4 = 29/

= 2 5/

≅ 2.4166667

4 0

2 years ago

One option in a roulette game is to bet $ 6

Eduardwww [97]

Answer:

<h2>We have 47% of chances to win those six bucks.</h2>

Step-by-step explanation:

In this problem we need to find the probabilty of winning thos $6.

According to the problem, we win $6 if we hit one red. We know that there are 18 red compartments, 18 black compartments and 2 neither red nor blakc, for a total of 38 compartments.

We need to use the standard probabilty definition which is the ratio between the number of events and the total number of outcomes. So,

$P=\frac{18}{38} \approx 0.47$

Which is equivalent to 47% of chances.

Therefore, we have 47% of chances to win those six bucks. (The Casino has the odds in their favor)

5 0

2 years ago