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Eva8 [605]
2 years ago
8

A 100 kg object is dropped from a height of 25 m. What is its velocity after falling ?

Mathematics
1 answer:
hjlf2 years ago
3 0

Answer:

the velocity would be 40.5

Step-by-step explanation:

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natima [27]
Could you provide just a little more info for this?
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I am totally stuck on this problem please help
Nonamiya [84]
V solid = π * 5^{2} * 13 / 3 + π * 5^{2} * 2 = 3.14 * 25 *13 / 3 + 3.14 * 50 = 340.16 + 157 = 497.16 ≈ 497.20 cubic centimeters;
3 0
4 years ago
2logbase4(x)-logbase4(5)=125
Goryan [66]
D:x>0\\\\ 2\log_4x-\log_45=125\\ \log_4x^2-\log_45=\log_44^{125}\\ \log_4\dfrac{x^2}{5}=\log_44^{125}\\ \dfrac{x^2}{5}=4^{125}\\ x^2=5\cdot4^{125}=5\cdot2^{250}=5\cdot(2^{125})^2\\ x=\sqrt{5\cdot2^{250}} \vee x=-\sqrt{5\cdot2^{250}}\\ x=\sqrt{5\cdot4^{125}} \vee x=-\sqrt{5\cdot4^{125}}\\ x=\sqrt{5\cdot(2^{125})^2} \vee x=-\sqrt{5\cdot(2^{125})^2}\\ x=\sqrt5 \cdot\sqrt{(2^{125})^2} \vee x=-\sqrt5 \cdot\sqrt{(2^{125})^2} \\ x=2^{125}\sqrt5 \vee x=-2^{125}\sqrt5

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Kruka [31]

Answer:

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Step-by-step explanation:

I hope this helps :D

5 0
3 years ago
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Kite EFGH is inscribed in a rectangle such that F and H are midpoints and EG is parallel to the side of the rectangle.
aivan3 [116]

Answer:

"The area of EFGH is always One-half of the area of the rectangle."

Step-by-step explanation:

<em>Graph is attached.</em>

<em />

The kite consists of 2 triangle, EFG and EHG.

The area of EFG:

\frac{1}{2}*EG*h

where h is the height from F to EG

The area of EHG:

\frac{1}{2}*EG*h_1

where h_1 is the height from H to EG

We also know that h + h _1 is the width of the rectangle and EG is the length of the rectangle

Thus,

Area of Kite = \frac{1}{2}*EG*h + \frac{1}{2}*EG*h_1 = RectangleLength*RectangleWidth

Also, Area of rectangle is rectangle length * rectangle width.

Thus, area of kite is always half of that of rectangle, the third choice is right.

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