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3 years ago

Solve the inequality 3(2x+7)>-3+5x.

Mathematics

1 answer:

lilavasa [31]3 years ago

7 0

Answer:

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Step-by-step explanation:

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A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the poli

pishuonlain [190]

Complete question:

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road.

Answer:

The red car is traveling along the road at 80.356 ft/s

Step-by-step explanation:

Given

Police car is 50 feet side off the road

Red car is 130 feet up the road

Distance between them is decreasing at the rate of 75 feet per sec

Let x be how far the police is off the road.

Let y be how far the red car is up the road.

Let h be the distance between the police and the red car.

This forms a right triangle so we can use the Pythagorean theorem, to solve for h

h² = x² + y²

h² = 50² + 130²

h² = 19400

h = √19400

h = 139.284 ft

Again;

Let dx/dt be how fast the police is traveling toward the road.

Let dy/dt be how fast the red car is traveling along the road.

Let dh/dt be how fast the distance between the police and the car is decreasing.

Recall that, h² = x² + y² (now differentiate with respect to time, t)

2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt)

divide through by 2

h(dh/dt) = x(dx/dt) + y(dy/dt)

since the police car is not, dx/dt = 0

and dy/dt is the how fast is the red car actually traveling along the road

139.284(75) = 50(0) + 130(dy/dt)

10446.3 = 0 + 130(dy/dt)

dy/dt = 10446.3 / 130

dy/dt = 80.356 ft/s

Therefore, the red car is traveling along the road at 80.356 ft/s

6 0

3 years ago

Drake wants to save $750 so that he can take a class on computer analysis for cars. The class is being held on various dates ove

STatiana [176]

C. Both options A and B will allow him to meet his goal.

Looking at Drake's situation after 4 weeks, he only has $470 saved. By his original plan, he should have had $500 saved. So he's $30 short of his goal and has 2 weeks until his originally planned class. If he goes with option A and takes the later class, he will save an additional $125 which is more than enough to make up the $30 short fall. So option A will work for him to save enough money for his class. With option B, he will save $140 for the last 2 weeks of his plan giving him a savings of $280 for the last 2 weeks. Adding the $470 he's already saved will give him a total savings of $470 + $280 = $750 which is enough for him to attend his class. So option B will also allow Drake to attend his desired class. Both options A and B allow him to meet his goal. Hence, the answer is "c".

8 0

3 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have evaluated the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

3 0

1 year ago

Arrange these numbers in increasing order: 216,000,000, 2.2 × 10 to the 3rd power, 3.1 × 10 to the 7th power, 310,000

ziro4ka [17]

216,000,000
2.2 × 10 to the 3rd power = 2.2 * 10^3 = 2,200
3.1 × 10 to the 7th power = 3.1 * 10^7 = 31,000,000
310,000

Increasing order:
2,200 , 310,000 , 216,000,000 , 31,000,000

So increasing order with original numbers:
2.2 × 10 to the 3rd power, 310,000, 216,000,000, 3.1 × 10 to the 7th power

7 0

3 years ago

Which Answer is a solution to the inequality?

marusya05 [52]

Answer:

The answer is D) (-1, -1)

Step-by-step explanation:

In order to find which answer is a solution, you have to put the ordered pair in for their given variables. If it produces a true statement, then it is a solution. In this case, D is the only one that produces a true statement.

2x + y > -4

2(-1) + (-1) > -4

-2 - 1 > -4

-3 > -4 (TRUE)

8 0

3 years ago

Solve the inequality 3(2x+7)>-3+5x.​

Solve the inequality 3(2x+7)>-3+5x.