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3 years ago

Pentagon ABCDE is shown on the coordinate plane below.

Mathematics

2 answers:

ANTONII [103]3 years ago

5 0

Point D will pass from (1,-2) to (-1,2)

The second choice is the correct one.

Flura [38]3 years ago

3 0

Answer:

Option 2 is correct. The ordered pair of D'(-1,2)

Step-by-step explanation:

If pentagon ABCDE is rotated 180° around origin to create pentagon A'B'C'D'E'

We need to find the ordered pair of point D'

We will use the formula of rotation of coordinate about origin.

$x'=x\cos\theta+y\sin\theta$

$y'=-x\sin\theta+y\cos\theta$

We need to find the coordinate of D' which is corresponding of D.

Coordinate of D: (1,-2)

$\text{Where, }x=1, y=-2\text{ and }\theta=180^{\circ}$

Now we find D':(x',y')

$x'=1\times \cos 180^{\circ}-2\times \sin 180^{\circ}=-1$

$y'=-1\times \sin 180^{\circ}-2\times \cos 180^{\circ}=2$

D' (-1,2)

Thus, The ordered pair of D'(-1,2)

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A radio arm saw has a circular cutting blade with a diameter of 10 inches and it spins at 2000rpm if there are 12 cutting teeth

FinnZ [79.3K]

Answer:

12565.11 teeth approximately cut cross the cutting surface per seconds.

Step-by-step explanation:

Since the diameter is 10inches, the circumference of the arm saw is 10π (circumference is πd where d is diameter).

The number of teeth on the cutting blade can be gotten by multiplying the diameter by 12 since we have 12 teeth per inch on the arm saw

Number of teeth on saw = 12 x 10π = 120π.

To get the number of teeth that cross the cutting surface in 2000rpm we multiple it by the total number of teeth in just one rpm.

So we have 2000 x 120π. But since we are asked to find it in revolution per seconds then 2000r/m = 2000r/60s ≈ 33.33

Therefore, we have

33.33 x 120π ≈ 12565.11 teeth

7 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:

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