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2 years ago

8

Determine if the expression 3c^5+7c^-1-c is a polynomial or not. If it is a polynomial, state the type and degree of the polynom

ial.

1 answer:

ludmilkaskok [199]2 years ago

3 0

Answer: It's not a polynomial

It's not a polynomial because the exponent of -1 in the second term. The exponents for any polynomial must be nonnegative whole numbers {0,1,2,3,...}

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The perimeter of the triangular park on the right is 16x-6. What is the missing length? The bottom side length is 5x+4 and the r

Romashka-Z-Leto [24]

Answer:

$(9x-5)\ units$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

The perimeter of a triangle is the sum of the length of their three sides

so

$P=a+b+c$

where

a,b,c are the length sides of the triangle

In this problem we have

$P=(16x-6)\ units\\a=(2x-5)\ units\\b=(5x+4)\ units$

substitute and solve for the missing length

$(16x-6)=(2x-5)+(5x+4)+c\\c=(16x-6)-(2x-5)-(5x+4)\\c=(9x-5)\ units$

3 0

3 years ago

Yolanda wanted to see if there was a connection between red hair and green eyes. she observed people walking past her on the str

loris [4]

Answer:

the answe is X= 14/ A :)

explanation: I took the exam review good luck!

5 0

2 years ago

Read 2 more answers

How do I simplify this exponent problem???

Nikolay [14]

Hello from MrBillDoesMath!

Answer:

(3/4) a^(-5)b^(-3)c^2

Discussion:

(18 a^-3b^2c^6)/ (24 a^2b^5c^4) =

(18/24) a^ (-3-2) b^(2-5) c^(6-4) =

as a^-3/a^-2 = a ^ (-3-2) = a^(-5), for examples

(3/4) a^(-5)b^(-3)c^2

Thank you,

MrB

8 0

2 years ago

A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)equals72 com

solmaris [256]

Answer:

Part (A)

1. Maximum revenue: $450,000

Part (B)

2. Maximum protit: $192,500

3. Production level: 2,300 television sets

4. Price: $185 per television set

Part (C)

5. Number of sets: 2,260 television sets.

6. Maximum profit: $183,800

7. Price: $187 per television set.

Explanation:

<u>0. Write the monthly cost and price-demand equations correctly:</u>

Cost:

$C(x)=72,000+70x$

Price-demand:

$p(x)=300-\dfrac{x}{20}$

Domain:

$0\leq x\leq 6000$

<em>1. Part (A) Find the maximum revenue</em>

Revenue = price × quantity

Revenue = R(x)

$R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x$

Simplify

$R(x)=300x-\dfrac{x^2}{20}$

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

$R'(x)=300-\dfrac{x}{10}$

Solve for R'(x)=0

$300-\dfrac{x}{10}=0$

$3000-x=0\\\\x=3000$

Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.

Hence, the maximum revenue is obtained when the production level is 3,000 units.

And it is calculated by subsituting x = 3,000 in the equation for R(x):

R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

Hence, the maximum revenue is $450,000

<em>2. Part (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. </em>

i) Profit(x) = Revenue(x) - Cost(x)

Profit (x) = R(x) - C(x)

$Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)$

$Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000$

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

Profit' (x) = -x/10 + 230

-x/10 + 230 = 0

-x + 2,300 = 0

x = 2,300

Thus, the production level that will realize the maximum profit is 2,300 units.

iii) Find the maximum profit.

You must substitute x = 2,300 into the equation for the profit:

Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

p(x) = 300 - x/20

p(2,300) = 300 - 2,300 / 20

p(2,300) = 185

Therefore, the company should charge a price os $185 for every television set.

<em>3. Part (C) If the government decides to tax the company $4 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?</em>

i) Now you must subtract the $4 tax for each television set, this is 4x from the profit equation.

The new profit equation will be:

Profit(x) = -x² / 20 + 230x - 4x - 72,000

Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

Profit'(x) = -x/10 + 226 = 0

-x/10 + 226 = 0

-x + 2,260 = 0

x = 2,260

Then, the new maximum profit is reached when the production level is 2,260 units.

iii) Find the maximum profit by substituting x = 2,260 into the profit equation:

Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000

Profit (2,260) = 183,800

Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800

iv) Find the price the company should charge for each set.

Substitute the number of units, 2,260, into the equation for the price:

p(2,260) = 300 - 2,260/20

p(2,260) = 187.

That is, the company should charge $187 per television set.

7 0

2 years ago

Wazi has sixty three 5 $ bills he takes the 5 $ bills to bank to exchange for 20$ how many 20$ bills does Wes I get

dmitriy555 [2]

15

63 x 5 = 315
315/20 = 15.75
hence, 15 $20 bills

4 0

3 years ago