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Nookie1986 [14]
3 years ago
15

Indicate in standard form the equation of the line passing through the given point and having the given slope.. . C(0, 4), m = 0

Mathematics
2 answers:
Lena [83]3 years ago
8 0
Well, in order to find it you just need to do this equation :

(y - 4)   = 0 (x - 0)

y - 4 = 0

y = 4

hope this helps
max2010maxim [7]3 years ago
7 0
 <span>standard form the equation of the line passing through the given point and having the given slope.. . C(0, 4), m = 0 </span>y-4 = 0(x-0) = 0 => y = 4 (this is a horizontal line) 
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Bogdan [553]

Answer:

21

Step-by-step explanation:

90+15=105

105÷5=21

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2 years ago
I need help fast
hram777 [196]

Answer:

On a coordinate plane, a line goes through (0, 3) and (2, 4) and another line goes through (0, 3) and (0.75, 0).

This answer almost coincide with option C. I suppose there was a mistype.

Step-by-step explanation:

The system of equations is formed by:

–x + 2y = 6

4x + y = 3

In the picture attached, the solution set is shown.

The first equation goes through (0, 3) and (2, 4), as can be checked by:

–(0) + 2(3) = 6

–(2) + 2(4) = 6

The second goes through (0, 3) and (0.75, 0), as can be checked by:

4(0) + (3) = 3

4(0.75) + (0) = 3

7 0
2 years ago
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Please help me 15 points :)
maw [93]

Answer:I don't really know sorry but I think its the second one

Step-by-step explanation:

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2 years ago
A physics student stands at the top of a hill that has an elevation of 56 meters. He throws a rock and it goes up into the air a
Virty [35]
round your answer to the nearest hundredth place the answer is a
3 0
3 years ago
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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