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3 years ago

Please help this ones pretty easy just not sure how to do it

Mathematics

1 answer:

BARSIC [14]3 years ago

3 0

Answer:

-8i + 51

Step-by-step explanation:

(3+5i)(2-9i)

6+10i-18i-45i²

**i² = -1.

6-8i + -45(-1)

= -8i + 51

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A new pencil is 19 centimeters long. Levi uses it for a month. The pencil now measures 8 centimeters.

pogonyaev

Answer:

11 cm

Step-by-step explanation:

Given:

Length of new pencil = 19 cm

Length of pencil after using a month = 8 cm

Question asked:

The pencil is centimeters shorter now than when it was new = ?

Solution:

Length of new pencil = 19 cm

Length of pencil after using a month = 8 cm

The pencil is centimeters shorter now than when it was new = 19 cm - 8 cm

= 11 cm

Length of pencil has been used during a month = 11 cm

8 0

3 years ago

Select the correct answer.

Nat2105 [25]

Answer:

Step-by-step explanation:

Both distances are in the scientific notation:

Earth - Sun = 9.3 * 10^7 miles

Saturn - Sun = 8.87 * 10^8 miles

8.87 * 10^8 - 9.3 * 10^7 =

= 88.7 *10^7 - 9.3 * 10^7 =

= 79.4 * 10^7 = 7.94 * 10 ^8 = 794,000,000 miles

Answer: Saturn is 7.94 * 10^8 miles farther from Sun than Earth is.

6 0

3 years ago

Planes X and Y are perpendicular. Points A, E, F, and G are points only in plane X. Points R and S are points in both planes X a

NemiM [27]

Given:

Planes X and Y are perpendicular to each other
Points A, E, F, and G are points only in plane X
Points R and S are points in both planes X and Y
Lines EA and FG are parallel

The lines which could be perpendicular to RS are EA and FG.

6 0

3 years ago

Read 2 more answers

Correct Answers only please!

Tcecarenko [31]

Answer:

C. $97

Step-by-step explanation:

The average of his wage for all 15 days is the sum of all wages for the 15 days divided by 15.

average wage for 15 days = (sum of wages for the 15 days)/15

The amount of wages during a number of days is the product of the average wage of those days and the number of days.

First 7 days:

average wage: $87

number of days: 7

total wages in first 7 days = 7 * $87/day = $609

Last 7 days:

average wage: $92

number of days: 7

total wages in last 7 days = 7 * $92/day = $644

8th day:

wages of the 8th day is unknown, so we let x = wages of the 8th day

total wages of 15 days = (wages of first 7 days) + (wages of 8th day) + (wages of last 7 days)

total wages of 15 days = 609 + x + 644 = x + 1253

average wage for 15 days = (sum of wages for the 15 days)/15

average wage for 15 days = (x + 1253)/15

We are told the average for the 15 days is $90/day.

(x + 1253)/15 = 90

Multiply both sides by 15.

x + 1253 = 1350

Subtract 1253 from both sides.

x = 97

Answer: $97

3 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.

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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

2 years ago