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3 years ago

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Alexander is blocking off several rooms in a hotel for guests coming to his wedding. The hotel can reserve small rooms that can

hold 3 people, and large rooms that can hold 4 people. Alexander reserved 4 more large rooms than small rooms, which altogether can accommodate 58 guests. Determine the number of small rooms reserved and the number of large rooms reserved.

1 answer:

romanna [79]3 years ago

3 0

Answer:

Small rooms=6

large rooms=10

Step-by-step explanation:

Let the number of small rooms be x

given alexander booked 4 more large rooms than the small rooms.

Therefore the number of large numbers be x+4.

Given each small room can accommodate 3 people.Therefore the total number of people in small rooms are 3x

Given each large room can accommodate 4 people. Therefore the total number of people in large rooms are 4(x+4)

Given total number of people accommodated=58

Total number of guests =number of guests in small rooms + number of guests in large rooms

$58=3x+4(x+4)$

$7x=42$

$x=6$

therefore the number of small rooms =6 and number of large rooms =10

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Part 1) Option C. Same side interior angles

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3 years ago

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Given <em>z</em> = 3 + <em>i</em>, right away we can find

(a) square

<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>

(b) modulus

|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(<em>z</em>) = arctan(1/3)

Then

<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))

<em>z</em> = √10 exp(<em>i</em> arctan(1/3))

or

<em>z</em> = √10 (cos(arctan(1/3)) + <em>i</em> sin(arctan(1/3))

(c) square root

Any complex number has 2 square roots. Using the polar form from part (d), we have

√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)

and

√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 2)

Then in standard rectangular form, we have

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and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that <em>z</em> lies in the first quadrant, so

0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2

which means

0 < 1/2 arctan(1/3) < <em>π</em>/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

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$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

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xeze [42]

Some parts are missing in the queston. Find attached the picture with the complete question

Answer:

$\large\boxed{\large\boxed{161}}$

Explanation:

Let's put the information in a table step-by step.

(number of remaining students)

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Condition

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At the end, there were 8 more seniors than juniors:

2/3×(S - 15) - 1/4×2(S - 15) = 8

Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

$2/3\times (S - 15) - 1/4\times 2(S - 15) = 8$

Mutilply all by 12:

$8(S - 15)-6(S - 15)=96$

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$8S-120-6S-90=96$

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$S=306/2=153$

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

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Just substitute S with 153 fo find the number of students that remained to perfom in the musical:

$2/3\times (153-15)+1/4\times 2(153-15)\\ \\ 2/3(138)+1/2(138)$

$161$

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