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3 years ago

A bank account appreciates at a rate if 0.05%

Mathematics

1 answer:

lidiya [134]3 years ago

6 0

Answer:

if this is a true or false question it would be false...

Step-by-step explanation:

each bank has different fees and percentage rates, hope this helps

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Could somebody solve this

Gre4nikov [31]

E^-3
e^-2
e^-1
1
e
e^2
e^3

7 0

3 years ago

Read 2 more answers

Noa hopes to pick the winning ticket during her math class’s raffle. the student who picks the ticket with a true statement will

Natali5045456 [20]

Answer: The winning ticket is $2(x-6)=-35$ is equivalent to $2x=-23$ .

Step-by-step explanation:

Let's check all the options

$4(x-5)=35$ is not equivalent to $4x=40$

$\text{since }4(x-5)=35\\\Rightarrow4x-20=35\\\Rightarrow4x=35+20\\\Rightarrow4x=55$

$8(x-7)=35$ is not equivalent to $8x=28$

$\text{since }8(x-7)=35\\\Rightarrow8x-56=35\\\Rightarrow8x=35+56\\\Rightarrow8x=91$

$2(x-6)=-35$ is equivalent to $2x=-23$

$\text{since }2(x-6)=-35\\\Rightarrow2x-12=-35\\\Rightarrow2x=-35+12\\\Rightarrow2x=-23$

$9(x-6)=-35$ is not equivalent to $9x=-29$

$\text{since }9(x-5)=-35\\\Rightarrow9x--40=-35\\\Rightarrow9x=-35+40\\\Rightarrow9x=5$

Thus, the winning ticket is $2(x-6)=-35$ is equivalent to $2x=-23$ .

4 0

3 years ago

Read 2 more answers

Alguien me puede ayudar en una tarea

Elden [556K]

Va :) de que es/que hay que hacer?

6 0

1 year 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

I will give BRAINLIEST to whoever is correct.

NikAS [45]

The correct answer is the last number line because with an inequality, the "equal to" sign below the "<" or ">" means the circle should be filled in. The line is traveling in a negative direction because the inequality says that n is less than -1.

5 0

3 years ago

Read 2 more answers