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Dmitry_Shevchenko [17]

2 years ago

14

Marsha gave the cashier $20 to pay for 3 pairs of socks the cashier gave her $5.03 in change. each pair of socks cost the same a

mount. what is the cost In dollars and cents for each pair of socks?

1 answer:

eduard2 years ago

5 0

Answer:

$4.99

Step-by-step explanation:

20-5.03=14.97

14.97/3=4.99

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During a manufacturing process, a metal part in a machine is exposed to varying temperature conditions. The manufacturer of the

mars1129 [50]

Answer:

Yes, it will reach or exceed 141 degree F

Step-by-step explanation:

Given equation that shows the temperature T in degrees Fahrenheit x minutes after the machine is put into operation is,

$T = 0.005x^2 + 0.45x + 125$

Suppose T = 141°F,

$\implies 141 = 0.005x^2 + 0.45x + 125$

$\implies 0.005x^2 + 0.45x + 125 - 141 =0$

$\implies 0.005x^2 + 0.45x - 16=0$

Since, a quadratic equation $ax^2 + bx + c =0$ has,

Real roots,

If Discriminant, $D = b^2 - 4ac \geq 0$

Imaginary roots,

If D < 0,

Since, $0.45^2 - 4\times 0.005\times -16 = 0.2025 + 32 > 0$

Thus, roots of -0.005x² + 0.45x + 125 are real.

Hence, the temperature can reach or exceed 141 degree F.

6 0

2 years ago

I need help yall. Brainliest for the first person who actually shows their work.<br> 30 pts btw

Oliga [24]

Answer:

1. 24.36

2. Do the same thing (follow the steps in the picture)

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Jill weighs 120 pounds and is gaining ten pounds each month. Paris weighs 150 pounds and is gaining 4 pounds each month. How man

lyudmila [28]

Answer:

5 months

Step-by-step explanation:

To solve this problem, we need to create and algebraic equation for when Jill's weight will be equal to Paris's. If we take their current weight and then add the additional weight gain per month times the number of months, we can calculate the amount of time it will take for their weight to be equal by solving for m.

$120+10m=150+4m\\10m-4m=150-120\\6m=30\\m=5$

8 0

2 years ago

Which measurement is most accurate to describe the width of a penny?

klasskru [66]

Answer:

19 mm

Step-by-step explanation:

A penny is not that large, and you can fit multiple pennies in the palm of your hand. This means that the only reasonable measurement is 19 mm, (& 8 cm, but that is only if you have really large hands). Therefore, (A) is your choice.

~

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:

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

General Formulas and Concepts:

<u>Calculus</u>

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:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

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

<u>Step 2: Find Limit</u>

Let's start out by <em>directly</em> 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 <em>evaluated</em> the given limit.

___

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

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

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

Unit: Limits

3 0

1 year ago