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AlexFokin [52]
2 years ago
7

The area on a wall covered by a rectangular poster is 294 square inches. The length of the poster is 1.5 times longer than the w

idth of the poster. What are the dimensions of the poster?
Mathematics
2 answers:
Dafna1 [17]2 years ago
7 0
A= l×w
296 = 1.5l × w

You should get 20 at the end
Alex787 [66]2 years ago
4 0
Hey there,

Your question states: <span>The area on a wall covered by a rectangular poster is 294 square inches. The length of the poster is 1.5 times longer than the width of the poster. What are the dimensions of the poster? </span>

l= 16 \ x  \ 1.25 = 20  in

Your correct answer would be \boxed{20}

Hope this helps.
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Carbon-141414 is an element which loses exactly half of its mass every 573057305730 years. The mass of a sample of carbon-141414
UkoKoshka [18]

Answer:

The function that models the mass of the carbon-14 sample remaining t years since the initial measurement is

M(t) = 741 e⁻⁰•⁰⁰⁰¹²¹ᵗ

with M(t) in grams and t in years.

Step-by-step explanation:

Radioactive reactions always follow a first order reaction dynamic

Let the initial concentration of Carbon-14 be M₀ and the concentration at any time be M

(dM/dt) = -kM (Minus sign because it's a rate of reduction)

(dM/dt) = -kM

(dM/M) = -kdt

 ∫ (dM/M) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from M₀ to M and the Right hand side from 0 to t.

We obtain

In (M/M₀) = -kt

(M/M₀) = e⁻ᵏᵗ

M(t) = M₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T) are related thus

T = (In2)/k

The half life is the time taken for the radioactive substance to decay to hAlf of its original amount, and according to the question, T = 5730 years

k = (In 2)/5730 = 0.000120968 /year. = 0.000121 /year

M(t) = M₀ e⁻ᵏᵗ

k = 0.000121 /year, M₀ = 741 grams

The equation then becomes

M(t) = 741 e⁻⁰•⁰⁰⁰¹²¹ᵗ

with M(t) in grams and t in years.

Hope this Helps!!!

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2 years ago
A professor surveys his students about their opinions on how well he taught the course. He puts this survey on the last page of
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Response/ NonResponse: Some students might be too tired to answer the question or in converse,  a student might feel that his or her grade might be reflected on the survey since it is not anonymous, he or she might sugarcoat the answer. (incorrect response bias)
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2 years ago
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aalyn [17]

Answer:

see the explanation

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of arc EH

Remember that the measure of the complete circle is equal to 360 degrees

so

arc\ EH+arc\ EF+arc\ FG+arc\ GH=360^o

Remember that

arc\ EF=m\angle EDF ----> by central angle

arc\ FG=m\angle FDG ----> by central angle

arc\ GH=m\angle GDH ---> by central angle

substitute the given values

arc\ EH+55^o+70^o+110^o=360^o

arc\ EH=360^o-235^o=125^o

step 2

Which arc is congruent to Arc E H?

we know that

arc\ EFG=arc\ EF+arc\ FG=55^o+70^o=125^o

therefore

arc EH is congruent with arc EFG

or

arc EH is congruent with minor arc GE

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2 years ago
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The amphitheater has two types of tickets available, reserved seats and lawn seats. The maximum capacity of the venue is 20,000
Tema [17]

Let the number of reserved tickets = x

Let the number of lawn seats = y

Constraint functions:

Maximum capacity means x+y\leq 20000

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Objective functions :

Maximum profit equation p = 65x +40y

Intersection points :

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p at (10000,10000) = 65(10000) + 40(10000) = $1050000

p at (20000,0) = 65(20000) + 40(0) = $1300000

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p at (5000,0) = 65(5000) + 40(0) =  $325000

Hence maximum profit occurs when all 20000 reserved seats are sold and the profit is $1300000

Please find attached the graph of it.

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