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Nimfa-mama [501]
3 years ago
7

In a theater, there are 4 long rows of chairs,each with an equal number of chairs There are 5 ahort rows of chairs,each with an

equal number of chairs On a crowded day, the ushers added 2 chairs to each long row and 3 chairs to each short row if x represents the number of chairs in each long row and y represents the number of chairs in each short row,the total number of chairs in the theater is given by the expression 2)Sly 3) An equivalent expression is
Mathematics
2 answers:
Nady [450]3 years ago
8 0

Answer:

An equivalent expression is 4x + 5y + 23

Step-by-step explanation:

There are 4 long rows of chairs and 3 short rows of chairs.

Let x be the number of chairs in long rows and y in short rows.

So total number of chairs in theater

C = 4x + 5y

Now on a crowded day 2 chairs were added in long rows so number of chairs in long rows becomes

= 4 (x + 2)

Similarly ushers add 3 chairs add in short rows so total number of chairs in short row = 5 ( y + 3 )

Now total chairs in theater

C' = 4 ( x + 2 ) + 5 ( y + 3 )

= 4x + 8 + 5y + 15

= 4x + 5y + 23

C' = C + 23   [Since C = ( 4x + 5y ) ]

Number of chairs added were 23 as compared to normal days.

Hoochie [10]3 years ago
7 0
<span>There are 4 long rows and 5 short rows in the theater. x represents the number of chairs in each long row and y represents the number of chairs in each short row.

So, total number of chairs in 4 long rows= 4x
Total number of chairs in 5 short rows = 5y
Total number of chairs in the theater on a normal day = 4x + 5y

When 2 chairs are added to each long row, the number of chairs will change to (x+2).
So, total number of chairs in 4 long rows will be = 4(x+2)

When 3 chairs are added to each short row, the number of chairs will change to (y+3)
So, total number of chairs in 5 short rows will be = 5(y+3)

Thus, total number of chairs in the theater in rush day = 4(x+2) + 5(y+3)
= 4x + 8 + 5y + 15
= 4x + 5y + 23

Thus we can say the number of chairs increase by 23 as compared to a normal day.
</span>
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Assume that y varies inversely with x. If y=1/3 when x=1/2, find y when x=1/4.
LenKa [72]

Answer:

y = \frac{2}{3}

Step-by-step explanation:

Given that y varies inversely with x then the equation relating them is

y = \frac{k}{x} ← k is the constant of variation

To find k use the condition y = \frac{1}{3} when x = \frac{1}{2} , thus

\frac{1}{3} = \frac{k}{\frac{1}{2} } = 2k ( divide both sides by 2 )

k = \frac{1}{6}

y = \frac{1}{6x} ← equation of variation

When x = \frac{1}{4} , then

y = \frac{1}{6(\frac{1}{4}) } = \frac{1}{\frac{3}{2} } = \frac{2}{3}

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3 years ago
What is the upper bound of the function f(x)=4x4−2x3+x−5?
inessss [21]

Answer:

(no global maxima found)

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = 4 x^4 - 2 x^3 + x - 5

Hint: | Global extrema of f(x) can occur only at the critical points or the endpoints of the domain.

Find the critical points of f(x):

Compute the critical points of 4 x^4 - 2 x^3 + x - 5

Hint: | To find critical points, find where f'(x) is zero or where f'(x) does not exist. First, find the derivative of 4 x^4 - 2 x^3 + x - 5.

To find all critical points, first compute f'(x):

d/( dx)(4 x^4 - 2 x^3 + x - 5) = 16 x^3 - 6 x^2 + 1:

f'(x) = 16 x^3 - 6 x^2 + 1

Hint: | Find where f'(x) is zero by solving 16 x^3 - 6 x^2 + 1 = 0.

Solving 16 x^3 - 6 x^2 + 1 = 0 yields x≈-0.303504:

x = -0.303504

Hint: | Find where f'(x) = 16 x^3 - 6 x^2 + 1 does not exist.

f'(x) exists everywhere:

16 x^3 - 6 x^2 + 1 exists everywhere

Hint: | Collect results.

The only critical point of 4 x^4 - 2 x^3 + x - 5 is at x = -0.303504:

x = -0.303504

Hint: | Determine the endpoints of the domain of f(x).

The domain of 4 x^4 - 2 x^3 + x - 5 is R:

The endpoints of R are x = -∞ and ∞

Hint: | Evaluate f(x) at the critical points and at the endpoints of the domain, taking limits if necessary.

Evaluate 4 x^4 - 2 x^3 + x - 5 at x = -∞, -0.303504 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-0.303504 | -5.21365

∞ | ∞

Hint: | Determine the largest and smallest values that f achieves at these points.

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-0.303504 | -5.21365 | global min

∞ | ∞ | global max

Hint: | Finally, remove the endpoints of the domain where f(x) is not defined.

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-0.303504 | -5.21365 | global min

Hint: | Summarize the results.

f(x) = 4 x^4 - 2 x^3 + x - 5 has one global minimum:

Answer: f(x) has a global minimum at x = -0.303504

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Answer:

\log_{13}(169)=2

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The given exponential form is:

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The number is 169.

In terms of logarithms, the base still remains the base and the exponent becomes the result of the logarithm of the number 169 to the given base 13.

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