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Veseljchak [2.6K]
3 years ago
11

Trapezoid WKLX has vertices W(2, −3), K(4, −3), L(5, −2) , and X(1, −2) . Trapezoid WKLX ​ is translated 4 units right and 3 uni

ts down to produce trapezoid trapezoid W'K'L'X' .
Which coordinates describe the vertices of the image?

W'(6, 0), K'(8, 0), L'(9, 1) , and X'(5, 1)
W′(6, −6), K′(8, −6), L′(9, −5) , and X′(5, −5)
W'(5, 1), K'(7, 1), L'(8, 2) , and X′(4, 2)
W′(−1, 1), K′(1, 1), L′(2, 2) , and X′(−2, 2)
Mathematics
2 answers:
guapka [62]3 years ago
7 0
Okay, so first of all, if the trapezoid was translated right 4,
Then all the (x)'s should be 4 more than the original value
For example: (2,-3) <span>→ (6,-3)
Now since the Trapezoid was translated down 3,
Then all the (y)'s should be 3 less than the original value
For example: (2,-3) </span><span>→ (2,-6)
Now do this to all the Vertices
</span>(6,-6)(8,-6)(9,-5)(5,-5)
Final Answer: <span>Option 2</span>
Vinvika [58]3 years ago
7 0
<span>W′(6, −6), K′(8, −6), L′(9, −5) , and X′(5, −5)

</span>make me the brainiest answer 
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\dfrac{2x-20}{3}=2x

Multiply both sides by 3:

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Subtract 2x from both sides:

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Divide both sides by 4:

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<span>11.34cm</span>
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f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans
Kruka [31]

Answer:

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

Step-by-step explanation:

We want to find the Riemann sum for \int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)

where \Delta{x}=\frac{b-a}{n}.

Step 1: Find \Delta{x}

We have that a=0, b=\frac{3\pi }{4}, n=6

Therefore, \Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}

Step 2: Divide the interval \left[0,\frac{3 \pi}{4}\right] into n = 6 sub-intervals of length \Delta{x}=\frac{\pi}{8}

a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b

Step 3: Evaluate the function at the left endpoints

f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3

f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386

f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964

f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527

f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0

f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527

Step 4: Apply the Left Riemann Sum formula

\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315

\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558

5 0
3 years ago
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