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ad-work [718]
3 years ago
10

Quartile of 20, 21, 25, 25

Mathematics
1 answer:
Fantom [35]3 years ago
3 0
Quartile or interquartile.
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A chain lying on the ground is 10 m long and its mass is 70 kg. How much work (in J) is required to raise one end of the chain t
DerKrebs [107]

Answer:

W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J

Step-by-step explanation:

Data Given: m = 70 kg , g = 9.8 ms^-2, h =10m.

For this case we can use the following formula:

W = \int_{x_i}^{x_f} F(x) dx

For this case we need to find an expression for the force in terms of the distance. And since on this case the total distance is 10 m long we can write the expression like this:

F(x) = \frac{ma}{10m}= \frac{mg}{10m} x

The only acceleration on this case is the gravity and if we replace the values given we got:

\frac{70 kg *9.8 m/s^2}{10m} x=68.6 x\frac{kg}{s^2}

Now we can find the required work with the following integral:

W= 68.6 \frac{kg}{s^2} \int_{0}^4 x dx

W= 34.3 \frac{kg}{s^2} x^2 \Big|_0^4

W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J

7 0
4 years ago
Sos HELP MEEEEE I HAVE NO CLUE PLZ EXPLAIN !!!​
OleMash [197]
2,376 square feet
**
Add the areas of all the sides of the shape.
A triangle's area: 1/2bh
The two triangles' area=24*9=216
The bottom side: 40*24=960
The two other sides: 30*40=1200
Add everything: 216+960+1200=2376
5 0
4 years ago
Evaluate h(5t^{2})given that h(t)=17-11t
Gala2k [10]
h(t)=17-11t
h(5t^{2})=17-11(5t^{2})
h(5t^{2})=17-55t^{2}
5 0
4 years ago
The full answer I don't understand
Slav-nsk [51]
DOB
that is from 55 to 145

acute means angle is less than 90 degrees
right means equal to 90
obtuse means greater than 90

so

55 to 145
145-55=90
it is right angle
4 0
4 years ago
Find vertex for f(x) = -x*2-2x-6. show work and sensable answer please.
frez [133]

Answer:

The vertex would be (-1, -5).

Step-by-step explanation:

In order to find this, first find the x-coordinate of the vertex. You can do this by calculating out -b/2a in which a is the coefficient of x^2 and b is the coefficient of x.

-b/2a

-(-2)/2(-1)

2/-2

-1

So we know the x-coordinate to be -1. Now we plug that into the equation and find the y value.

-x^2 - 2x - 6

-(-1)^2 - 2(-1) - 6

-(1) + 2 - 6

-5

4 0
3 years ago
Read 2 more answers
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