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3 years ago

What’s the answer to number 1 ? Show how you solved please !!!!

Mathematics

1 answer:

Elodia [21]3 years ago

5 0

Dan had 52
shown work: 29 x 2=58 58-6=52

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<img src="https://tex.z-dn.net/?f=%5Csqrt%7B0.81%7D" id="TexFormula1" title="\sqrt{0.81}" alt="\sqrt{0.81}" align="absmiddle" cl

Mandarinka [93]

Answer: .9

Step-by-step explanation:

3 0

3 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

2 years ago

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost and same-day tickets cost . Fo

AveGali [126]

Answer:

83746+4747+48484=3943848343

Step-by-step explanation:

Easy

8 0

2 years ago

Find the exact location of all the relative and absolute extrema of the function. HINT [See Examples 1 and 2.] (Order your answe

icang [17]

Answer:

(-1, -32) absolute minimum
(0, 0) relative maximum
(2, -32) absolute minimum
(+∞, +∞) absolute maximum (or "no absolute maximum")

Step-by-step explanation:

There will be extremes at the ends of the domain interval, and at turning points where the first derivative is zero.

The derivative is ...

h'(t) = 24t^2 -48t = 24t(t -2)

This has zeros at t=0 and t=2, so that is where extremes will be located.

We can determine relative and absolute extrema by evaluating the function at the interval ends and at the turning points.

h(-1) = 8(-1)²(-1-3) = -32

h(0) = 8(0)(0-3) = 0

h(2) = 8(2²)(2 -3) = -32

h(∞) = 8(∞)³ = ∞

The absolute minimum is -32, found at t=-1 and at t=2. The absolute maximum is ∞, found at t→∞. The relative maximum is 0, found at t=0.

The extrema are ...

(-1, -32) absolute minimum
(0, 0) relative maximum
(2, -32) absolute minimum
(+∞, +∞) absolute maximum

_____

Normally, we would not list (∞, ∞) as being an absolute maximum, because it is not a specific value at a specific point. Rather, we might say there is no absolute maximum.

5 0

3 years ago

Which number line shows the solution set to this inequality? -2x + 9 < x − 9

kotykmax [81]

Answer:

see below

Step-by-step explanation:

-2x + 9 < x − 9

Add 2x to each side

2x-2x + 9 < 2x+x − 9

9 < 3x-9

Add 9 to each side

9+9 < 3x

18 < 3x

Divide each side by 3

18/3 < 3x/3

6 < x

Open circle at 6 line going to the right

5 0

3 years ago