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garik1379 [7]
4 years ago
5

If I make a 40.2 on a test what fraction of the questions did I answer correctly

Mathematics
1 answer:
Kruka [31]4 years ago
6 0

Answer:

You answered 201/500 of the questions correctly

Step-by-step explanation:

40.2/100=402/1000=201/500

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B

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The area of a rectangle court is 433.37 Square meters and the length of the court is 28.7 meters . What is width of the court?
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w=15.1m

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the thrid one

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5 0
3 years ago
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Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Only 1 try
vagabundo [1.1K]

Using the shell method, the volume is

\displaystyle 2\pi \int_0^1 (2-x) \cdot 8x^3 \, dx = 16\pi \int_0^1 (2x^3 - x^4) \, dx

Each cylindrical shell has radius 2-x (the horizontal distance from the axis of revolution to the curve y=8x^3); has height 8x^3 (the vertical distance between a point on the x-axis in 0\le x\le1 and the curve y=8x^3).

Compute the integral.

\displaystyle 16 \pi \int_0^1 (2x^3 - x^4) \, dx = 16\pi \left(\frac{x^4}2 - \frac{x^5}5\right) \bigg|_{x=0}^{x=1} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 16\pi \left(\frac12 - \frac15\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{24}5\pi = \boxed{4.8\pi}

6 0
2 years ago
We are given AB ∥ DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal
ohaa [14]

Answer: AAA similarity.


Step-by-step explanation:  CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have

∠CED ≅ ∠CBA,

∠CDE ≅ ∠CAB

and

∠DCE ≅ ∠ACB [same angle]

Hence, by AAA (angle-angle-angle) similarity,

△CED ~ △ABC.

Thus, the correct option is AAA similarity.


8 0
3 years ago
Read 2 more answers
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