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3 years ago

What is 80 x 40? Make 5 paragraphs stating the answer.

1 answer:

7 0

80 x 40 is 320. I know 80 x 40 is 320 because 80 + 80 + 80 + 80 is 320. I also know 80 x 40 is 320 because 8 x 4 is 36. If you pretend the 0 is not there it’s 8 x 4. I know you can get the answer quicker that way because it’s just times tables.

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Write an equation in slope-intercept form for the line with slope 4 and y-intercept - 2

ladessa [460]

Answer:

The slope intercept form is typically in the form of

y=mx+c

c=-2

m=4

The slope intercept form of the equation is

y=4x-2

6 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

Examine the system of equations. y = 3 2 x − 6, y = −9 2 x + 21 Use substitution to solve the system of equations. What is the v

nadezda [96]

Answer:

Step-by-step explanation:

Given the system of equations y = 3/2 x − 6, y = −9/2 x + 21, since both expressions are functions of y, we will equate both of them to find the variable x;

3/2 x − 6 = −9/2 x + 21,

Cross multiplying;

3(2x+21) = -9(2x-6)

6x+63 = -18x+54

collecting the like terms;

6x+18x = 54-63

24x = -9

x = -9/24

x = -3/8

To get the value of y, we will substitute x = -3/8 into any of the given equation. Using the first equation;

y = 3/2x-6

y = 3/{2(-3/8)-6}

y = 3/{(-3/4-6)}

y = 3/{(-3-24)/4}

y = 3/(-27/4)

y = 3 * -4/27

y = -4/9

Hence, the value of y is -4/9

4 0

3 years ago

Read 2 more answers

Simplify.

DIA [1.3K]

Hey, do you need to simplify all of them?

8 0

3 years ago

Solve 41-9+(8×2.3)-15+(2.1×4)

dsp73

(8x2.3)= (18.4)
(2.1x4)= (8.4)
Now you should have:
41-9+18.4-15+8.4
9+18= 27
15+8.4= 23.4
Should have:
41-27-23.4= -9.4

6 0

3 years ago