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polet
8 days ago
10

An equiangular triangle has one side of length six inches. What is the height of the triangle, drawn from that side, to the near

est tenth of an inch?

Mathematics
1 answer:
AnnZ [11.9K]8 days ago
5 0
The height of the triangle measures 5.2 inches. An equiangular triangle has equal measures for all interior angles, which means it is equivalent to an equilateral triangle. Since all angles equal 60 degrees, applying the Pythagorean theorem enables us to calculate the height based on the provided side length.
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A 1500 kg hippo is completely submerged, standing on the bottom of a lake. What is the approximate value of the upward normal fo
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Answer:

Aproximadamente 428 N

Step-by-step explanation:

Peso = 1,500 * 9.8 = 14,700 N

Densidad = Masa ÷ Volumen

1,030 = 1,500 ÷ V

V = 1,500 ÷ 1,030 = 1.46 m^3.

La fuerza de flotación = Densidad * g * V

La fuerza de flotación = 1,000 * 9.8 * (1,500 ÷ 1,030)

La fuerza de flotación = 9,800 * (1,500 ÷ 1,030) = 14,272 N.

La fuerza neta = 14,700 – [(9,800 * (1,500 ÷ 1,030)]

7 0
1 month ago
Select the sequence of transformations that will carry rectangle A onto rectangle A'. A) reflect over y-axis, rotate 90° clockwi
PIT_PIT [11953]

Answer:

Each of the 4 arrangements will produce a rectangle.

Explanation:

Transforming a rectangle through rotation or translation will not alter its rectangular shape. This principle also applies when reflecting it across any axis. Thus, every sequence among the four provided will result in a rectangle.

6 0
1 month ago
Read 2 more answers
Hi! I'm confused with this question, It's about tessellations. Can someone give a complete example of a possible answer? It woul
AnnZ [11980]
Yes, this reflects the correct answer as [[TAG_10]]
7 0
14 days ago
Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab
Svet_ta [12349]

Response:  a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Detailed explanation:

In Problem 8-4, the computer time-sharing system experiences teleport inquiries at an average rate of 0.1 per millisecond. We are tasked with determining the probabilities of the inquiries over a specific period of 50 milliseconds:

Given that

\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5

Applying the Poisson process, we find that

(a) at most 12

probability=  P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980

(b) exactly 13

probability= P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013

(c) more than 12

probability= P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020

(d) exactly 20

probability= P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026

(e) within the range of 10 to 15, inclusive

probability=P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318

Thus, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0
1 month ago
The general form of the equation of a circle is x2 + y2 + 8x + 22y + 37 = 0. The equation of this circle in standard form is (x
Leona [12178]
Greetings: 
<span>x² + y² + 8x + 22y + 37 = 0
(x² +8x) +(y² +22y) +37 = 0

</span>(x² +8x+4²)-4² +(y² +22y+11²) -11²+37 = 0
(x+4)² +(y+11)²-16-121+37 =0

(x+4)² +(y+11)² =10²...(<span>standard form )
</span><span>The circle's center is located at (-4, -11) and has a radius of 10</span>
3 0
24 days ago
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