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3 years ago

Best/correct answer gets brainliest!!

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

3 0

Answer:

Same side interior angles of parallel lines are supplementary.

Step-by-step explanation:

5a + 10 + 3a + 10 = 180; 8a + 20 = 180; 8a = 160; a = 20; m<4 = 5xz + 10 = 5 * 20 + 10 = 100 + 10 = 110.

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The cost of a car rental is $25 per day plus 22 per mile. You are on a daily budget of $90. Write and solve an inequality to f

lara31 [8.8K]

Answer: You can drive, at most, 2.95 miles per day.

x ≤ 2.95

Step-by-step explanation:

In a single day, you can spend at most $90

Then if C represents the cost of renting the car, then we will have the inequality:

C ≤ $90

Now let's find the equation for C.

We know that we have a fixed cost of $25 plus $22 per mile, then if you drive x miles, the total cost will be $25 plus x times $22, or:

C = $25 + x*$22.

We can now replace that in the inequality:

$25 + x*$22 ≤ $90

Now let's isolate the variable x

x*$22 ≤ $90 - $25

x*$22 ≤ $65

x ≤ $65/$22 = 2.95

x ≤ 2.95

You can drive at most, 2.95 miles per day.

To find other inequalities with the same solution we can start with the solution:

x ≤ 2.95

Now let's multiply both sides by a number (the units of the number can be dollars, in that way we can make a similar problem)

Let's multiply both sides by $10:

x*$10 ≤ 2.95*$10 = $29.5

x*$10 ≤ $29.5

Now let's add the same number in both sides, for example, $5.

x*$10 + $5 ≤ $29.5 + $5 = $34.5

x*$10 + $5 ≤ $34.5

We could write this problem as:

"To rent a cab in your city, you have an initial cost of $5, plus $10 for each mile driven. How many miles could you drive if at most you can spend $34.50?"

You could be more creative with the problem, but this is the way in which you can craft problems of this type when you already know the solution.

5 0

3 years ago

What is the answer to the question?

Mumz [18]

Answer:

The answer is 4 zeros

Step-by-step explanation:

(6×5)×10³

30×1000=30000

Therefore 4 zeros in 30000

8 0

4 years ago

Read 2 more answers

The following triangle. check all that apply

Amiraneli [1.4K]

Obtuse and isosceles

Isosceles means that at least two of the sides are congruent or equal.

Obtuse means that one angle of the triangle is greater than 90 degrees.

So, your answer is D and E

:)))

4 0

3 years ago

Read 2 more answers

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:

0. Write the monthly cost and price-demand equations correctly:

Cost:

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

Price-demand:

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

Domain:

$0\leq x\leq 6000$

1. Part (A) Find the maximum revenue

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

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. 

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.

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?

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

3 years ago

Diameter of a circle is two units. What is the radius of the circle?

bonufazy [111]

Answer is 3.5 feet enjoy

5 0

3 years ago