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2 years ago

Rex puts $5,000 in a savings account with a simple

Mathematics

1 answer:

Mashcka [7]2 years ago

5 0

Answer:

Rex earns $250 in interest after 2 years.

Step-by-step explanation:

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Can I Get A Little Help Here?

Musya8 [376]

Hey there :)

We know the formula for simple interest is
I = prt
{ The picture includes what each stands for }

Lehena:
Initial amount = $100
Interest rate = 2.5% = $\frac{2.5}{100} = 0.025$
Time = 3 years
I = ( 100 )( 0.025 )( 3 )
= 7.5$

Total = $100 + $7.5 = $107.5

Marty:
Initial amount = $100
Interest rate = 2% = $\frac{2}{100} = 0.02$
Time = 3 years
I = ( 100 )( 0.02 )( 3 )
= $6

Total = $100 + $6 = $106

How much more than Marty did Lehena receive?
= Total Amount of Lehena - Total Amount of Marty
= 107.5 - 106
= $1.5

Your answer will be option B) $1.50

7 0

2 years ago

Es of two children are 11 and 8 years.

nikdorinn [45]

Answer:

In 5 years the product of their ages will be 208

Step-by-step explanation:

The age of two children is 11 and 8 years.

Let's call x the number of years ahead.

We need to find when the product of their future ages is 208. The 11 years old child will be 11+x years old and the other child will be 8+x years, thus:

(11+x)(8+x)=208

Operating:

$88+11x+8x+x^2=208$

Simplifying:

$x^2+19x-120=0$

Factoring:

(x-5)(x+24)=0

Solving:

x=5, x=-24

The negative solution is not valid, thus x=5

In 5 years the product of their ages will be 208

3 0

1 year ago

Write the equation of a line perpendicular to 5x - 7y =-6 that passes through the point (-5,6)

Likurg_2 [28]

9514 1404 393

Answer:

7x +5y = -5

Step-by-step explanation:

You can find the perpendicular line by swapping the x- and y-coefficients, and negating one of them. The new constant can be found by using the given point.

7x +5y = 7(-5) +5(6) = -5

The perpendicular line through (-5, 6) is ...

7x +5y = -5

3 0

2 years ago

Please help me thanks if u do

aniked [119]

Answer: a

Step-by-step explanation: look at the chart

3 0

2 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

1 year ago