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FinnZ [79.3K]
3 years ago
14

HELP ME PLEASE!

Mathematics
1 answer:
Delicious77 [7]3 years ago
7 0
Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides).  If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.

Let's see whether they do.

8^2 + 8^2 = 11^2  ???
 64    + 64 = 121?  NO.  This is not a right triangle.

If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:

8^2 + 8^2 = d^2, where d=hypotenuse.  Then 64+64 = d^2, and 

d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.

11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.
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A rectangular tank is 39feet long, 25 feet wide 18 feet high there is 11 feet of water in the tank if a valve is open at the bot
Aliun [14]

Answer:

1596.2 ft/m

Step-by-step explanation:

<u>Data</u>

  • height of water, h = 11 ft
  • acceleration of gravity, g = 32.17 ft/s²

Given that 60 seconds are equivalent to 1 minute:

g = 32.17 ft/s² * 60² s²/1 min² = 115812 ft/min²

From Bernoulli's theorem:

velocity = √(2gh)

velocity = √(2*115812*11)

velocity = 1596.2 ft/m

5 0
3 years ago
Solve for x. <br> bx-10=8
gayaneshka [121]

Answer: x = 18/b

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3 0
3 years ago
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Solve the following equation: (Find x) <br>a) (x-5)²+(2x-10)(x+3)-x²+25​
Alex73 [517]

Answer:

hence the value of x is either 5 or 2

x=5 or x=2

4 0
2 years ago
What are 4 similarities between old imperialism and new imperialism?
snow_tiger [21]

Answer:

Old Imperialism led to exploration of new trade routes, establishment of new settlements in new lands and ultimately led to establishment of Political rule in those lands. Under New Imperialism, Nations ruled smaller colonial areas.

Step-by-step explanation:

5 0
3 years ago
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If an arrow is shot upward on Mars with a speed of 62 m/s, its height in meters t seconds later is given by y = 62t − 1.86t². (R
Soloha48 [4]

Answer:

Approximately 58.28\; \rm m \cdot s^{-1}.

Step-by-step explanation:

The velocity of an object is the rate at which its position changes. In other words, the velocity of an object is equal to the first derivative of its position, with respect to time.

Note that the arrow here is launched upwards. (Assume that the effect of wind on Mars is negligible.) There would be motion in the horizontal direction. The horizontal position of this arrow will stays the same. On the other hand, the vertical position of this arrow is the same as its height: y = 62\, t - 1.86\, t^2.

Apply the power rule to find the first derivative of this y with respect to time t.

By the power rule:

  • the first derivative of t (same as
  • the first derivative of t^2 (same as t to the second power) with respect to

Therefore:

\begin{aligned}\frac{dy}{d t} &= \frac{d}{d t}\left[62 \, t - 1.86\, t^2\right] \\ &= 62\,\left(\frac{d}{d t}\left[t\right]\right) - 1.86\, \left(\frac{d}{d t}\left[t^2\right]\right) \\ &= 62 \times 1 - 1.86\times\left(2\, t) = 62 - 3.72\, t\end{aligned}.

In other words, the (vertical) velocity of this arrow at time t would be (62 - 3.72\, t) meters per second.

Evaluate this expression for t = 1 to find the (vertical) velocity of this arrow at that moment: 62 - 3.72 \times 1 =58.28.

6 0
3 years ago
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