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sp2606 [1]
3 years ago
14

Is a kilogram lighter or heavier than a gram

Mathematics
1 answer:
skad [1K]3 years ago
6 0
A kilogram is equal to 1000 grams. This means that 1000 grams (1 kilogram) is heavier than 1 gram.
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What is the value of x
Anna11 [10]

Answer:

28

Step-by-step explanation:

14x2=28

Therefore x=28

8 0
3 years ago
What is the domain and range of g(x)= -|x+2| -1
solmaris [256]

g(x)= -|x+2| -1


The domain is the input values, or x values

it is all real number


The range is the output values, or g(x)

the absolute value at is smallest is 0

0-1 is -1  this is the max since the absolute value is multiplied by -1

g(x) <=-1

the output is less than or equal to -1

4 0
3 years ago
D'Quans grandmother made a quilt for his bed.the quilt is 2.44 meters long and 1.83 meters wide.what is the area of the quilt in
Ede4ka [16]
The best way to do this is to first change each of the measurements into feet and then calculate the area:

2.44m = 8.01ft
1.83m = 6.00ft

When multiplied together (to give the area) we get:

8.01*6.00 = 48.1ft²
5 0
3 years ago
Read 2 more answers
The value of a $25,000 car depreciates at a rate of 12% per year. What will the car be worth in 5 years?
Scorpion4ik [409]

The formula for depreciation is:

Value = Starting value x (1 - rate)^time

Using the given values:

Value = 25,000 x (1-0.12)^5

Value = 25,000 x 0.88^5

Value = $13,193.30 (Round the answer as needed)

3 0
3 years ago
Read 2 more answers
a) find center of mass of a solid of constant density bounded below by the paraboloid z=x^2+y^2 and above by the plane z=4.(b) F
slava [35]

Answer: x-bar = y-bar = 0 whereas z-bar = 8/3

               M= (c^2)/8 which is intern equal to 2\sqrt{2}

Step-by-step explanation:

           Find the area, by setting the limits as

               = 4\cdot \int _0^{\frac{\pi }{2}}\int _0^2\int _{r^2}^4\:rdzdrd\theta

                =4\cdot \int _0^{\frac{\pi }{2}}\int _0^2r\cdot \:4-r^3drd\theta

                =4\cdot \int _0^{\frac{\pi }{2}}4d\theta

                 =8\pi

Therefore;

Mxy=  \int _0^{2\pi }\int _0^2\int _{r^2}^4\:zrdzdrd\theta

       z-bar = 8/3

M= 8\pi dividing it into two volume gives us = 4\pi

means  4\pi =\int _0^{2\pi }\int _0^{\sqrt{c}}\int _r^c\:rdzdrd\theta

             4\pi =\left(\pi c^2-2\pi \frac{c^{\frac{3}{2}}}{3}\right)

              c=2\sqrt{2}

       

                 

                 

5 0
3 years ago
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