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krok68 [10]
3 years ago
15

How much does the Standard suite cost per night?

Mathematics
2 answers:
Amanda [17]3 years ago
7 0

Answer:

  B) $77.79

Step-by-step explanation:

I just took the test

kotykmax [81]3 years ago
3 0
B is the answer i believe.
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The Lakewood Wildcats won 5 out of their first seven games in this year. There are 28 games in the season. About how many games
allochka39001 [22]

Answer:

The total number of matches expected to be won by Lakewood Wildcats in this season  is 20.

Step-by-step explanation:

The number of games played by Lakewood Wildcats = 7

Number of matches won by Lakewood = 5

or, the ratio of Won : Played = 5: 7

Total numbers of games in the season = 28

Let the Lakewood Wildcats win m number of games.

Here, ratio of Won Matches : Played Matches = m : 28

Now, by RATIO OF PROPORTIONALITY:

\frac{5}{7}   = \frac{m}{28}

⇒m = \frac{28 \times 5}{7}  = 20

or, m = 20

Hence, the total number of matches expected to be won by Lakewood Wildcats out of total 28 matches is 20.

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3 years ago
How many 4 inch by 8 inch bricks are needed to build a walk 6 feet wide and 24 feet long
spayn [35]
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3 years ago
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Answer

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Step-by-step explanation:

7 0
3 years ago
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Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

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3 years ago
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What is the answer to 1.97 x 10^3 =
Gelneren [198K]
<h2>Answer:</h2><h2>1970</h2><h2 /><h2>Hope this helps!!</h2>

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