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Lapatulllka [165]

3 years ago

15

Michelle found a new violin on sale at 30% off. how much would she pay the cashier if it originally sells for $250 in a city tha

t had no sales tax

1 answer:

Pavlova-9 [17]3 years ago

5 0

100%-30% = 70%

70%=0.70

250*0.70 = 175

she would pay $175

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Help with num 9 please. thanks

Alik [6]

Answer:

See Below.

Step-by-step explanation:

We want to show that the function:

$f(x) = e^x - e^{-x}$

Increases for all values of <em>x</em>.

A function is increasing whenever its derivative is positive.

So, find the derivative of our function:

$\displaystyle f'(x) = \frac{d}{dx}\left[e^x - e^{-x}\right]$

Differentiate:

$\displaystyle f'(x) = e^x - (-e^{-x})$

Simplify:

$f'(x) = e^x+e^{-x}$

Since eˣ is always greater than zero and e⁻ˣ is also always greater than zero, f'(x) is always positive. Hence, the original function increases for all values of <em>x.</em>

8 0

3 years ago

How do you simplify 2(m+5)+8(6m+1)

galina1969 [7]

You must use distributive property to get 2m+10+48m+8 and then add all like terms. You get 50m+18 as your final answer.

5 0

3 years ago

Read 2 more answers

Law Incorporation [45]

Based on her results, if she flipped the coins another 50 times, she should expect to flip heads 20 times. If Jessica flips a coin 100 times and gets 40 times heads and the next time she flips a coin 50 times she should get 20 because, 100 divide 2 is 50 so you would have to divide 40 with 2 and get the answer of 20.

3 0

3 years ago

Find the vertical and horizontal asymptotes, domain, range, and roots of f (x) = -1 / x-3 +2.

gladu [14]

Answer:

Vertical asymptote: $x=3$

Horizontal asymptote: $f(x) =2$

Domain of f(x) is all real numbers except 3.

Range of f(x) is all real numbers except 2.

Step-by-step explanation:

Given:

Function:

$f (x) = -\dfrac{1 }{ x-3} +2$

One root, $x = 3.5$

To find:

Vertical and horizontal asymptote, domain, range and roots of f(x).

Solution:

First of all, let us find the roots of f(x).

<em>Roots of f(x) means the value of x where f(x) = 0</em>

$0= -\dfrac{1 }{ x-3} +2\\\Rightarrow 2= \dfrac{1 }{ x-3}\\\Rightarrow 2x-2 \times 3=1\\\Rightarrow 2x=7\\\Rightarrow x = 3.5$

One root, $x = 3.5$

Domain of f(x) i.e. the values that we give as input to the function and there is a value of f(x) defined for it.

For x = 3, the value of f(x) $\rightarrow \infty$

For all, other values of $x$ , $f(x)$ is defined.

Hence, Domain of f(x) is all real numbers except 3.

Range of f(x) i.e. the values that are possible output of the function.

f(x) = 2 is not possible in this case because something is subtracted from 2. That something is $\frac{1}{x-3}$ .

Hence, Range of f(x) is all real numbers except 2.

Vertical Asymptote is the value of x, where value of f(x) $\rightarrow \infty$ .

$-\dfrac{1 }{ x-3} +2 \rightarrow \infty$

It is possible only when

$x-3=0\\\Rightarrow x=3$

$\therefore$ vertical asymptote: $x=3$

Horizontal Asymptote is the value of f(x) , where value of x $\rightarrow \infty$ .

$x\rightarrow \infty \Rightarrow \dfrac{1 }{ x-3} \rightarrow 0\\\therefore f(x) =-0+2 \\\Rightarrow f(x) =2$

$\therefore$ Horizontal asymptote: $f(x) =2$

Please refer to the graph of given function as shown in the attached image.

5 0

3 years ago

Verify that this trigonometric equation is an identity?

Vlada [557]

$\cot x\sec^4 x=\cot x+2\tan x+\tan^3x\\\\L=\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\cos^4x}=\dfrac{1}{\sin x}\cdot\dfrac{1}{\cos^3x}=\dfrac{1}{\sin x\cos^3x}\\\\R=\dfrac{\cos x}{\sin x}+2\cdot\dfrac{\sin x}{\cos x}+\dfrac{\sin^3x}{\cos^3x}\\\\=\dfrac{\cos x\cos^3x}{\sin x\cos^3x}+\dfrac{2\sin x\cos^2x}{\cos x\sin x\cos^2x}+\dfrac{\sin^3x\sin x}{\cos^3x\sin x}\\\\=\dfrac{\cos^4x+2\sin^2x\cos^2x+\sin^4x}{\sin x\cos^3x}\\\\=\dfrac{(\cos^2x)^2+2\sin^2x\cos^2x+(\sin^2x)^2}{\sin x\cos^3x}$

$=\dfrac{(\cos^2x+\sin^2x)^2}{\sin x\cos^3x}=\dfrac{1}{\sin x\cos^3x}=L\\\\Used:\\\tan(a)=\dfrac{\sin(a)}{\cos(a)}\\\cot(a)=\dfrac{\cos(a)}{\sin(a)}\\\sec(a)=\dfrac{1}{\cos(a)}\\\sin^2a+\cos^2a=1\\(a+b)^2=a^2+2ab+b^2$

8 0

3 years ago