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2 years ago

What is the interest rate if the amount borrowed is $11,000 for 6 years and the interest paid is $1800?

1 answer:

4 0

The formula for simple interest is ...
I = Prt
1800 = 11000*r*6
1800/66000 = r = 0.02727... (repeating 27)

The simple interest rate is approximately 2.73%.

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Luda [366]

The answer is -4 final answer

3 0

2 years ago

Please help with number 1

bezimeni [28]

Answer:

19.0681

Step-by-step explanation:

Given in the question that,

angle from ted to the dog = 60° with the ground

height of ted from the ground = 16ft

To find,

distance between dog and the door of ted's building

Considering the scenario make a right angle triangle:

<h3>By using pythagorus theorem:</h3>

Tan 40 = opposite / adjacent

Tan 40 = height / distance between dog and the door

Tan 40 = 16ft / x

x = 16 / tan40

x = 19.068057

x ≈ 19.0681 (nearest to thousand)

So, the dog need to walk 19.0681ft to reach the open door directly below Ted.

7 0

2 years ago

Now go to diagonals of parallelograms. You'll see two line segments, and marked with their midpoints, E and F. Verify that E and

AURORKA [14]

Answer:

AE=4

EC=4

BF=2.24

FD=2.24

Step-by-step explanation:

here we go honey

5 0

2 years ago

What is 15 1/2 minus 11 3/8?

Anna35 [415]

First change the 1/2 to eighths. multiply numerator and denominator by 4.now subtract 4 1/8

3 0

2 years ago

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$

8 0

2 years ago