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xeze [42]
4 years ago
7

Find the sum. 1/4 + 2/7 a. 9/14 b. 3/11 c. 15/28 d. 3/28

Mathematics
1 answer:
Dmitry_Shevchenko [17]4 years ago
5 0
1/4 + 2/7 =

The denominators 4 and 7 lcm are 28

So,
1/28 + 2/28 = 3/28

3/28 is the answer to your question.
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13 A cylindrical iron rod 8 cm high and 6 cm in diameter stands in a cylindrical tin 12 cm in diameter. See Fig. 15.22. Water is
Eduardwww [97]

Answer:

2cm

Step-by-step explanation:

The volume of water displaced by the rod is

v = πr^2 h = π * 3^2 * 8 = 72π cm^3

The drop in water level is the rod volume divided by the cross-section area of the tin, or

72π/(π*6^2) = 2 cm

4 0
3 years ago
Read 2 more answers
6th grade math help me plzzzz
Ira Lisetskai [31]

Answer:

n=-3.1

Step-by-step explanation:

To solve this, first subtract 3 from both sides of the equation to get 2n=-6.2. Next, divide both sides of the equation by 2 n order to get the final answer of n=-3.1

4 0
3 years ago
Nikki spent 2/5 of her birthday money on new shoes and 1/4 of her birthday money on a new headband.
quester [9]

Answer:

/20

Step-by-step explanation

7 0
2 years ago
If the circle has the same diameter as the edge length of the square, then the area of this circle is ___________the area of the
WINSTONCH [101]

Answer:

The area of this circle is (\frac{\pi}{2} )  the area of the square.

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Therefore, Φsquare is (\frac{2}{\pi} ) ϕcircle

Step-by-step explanation:

Area of the circle is given by;

A_c = \frac{\pi d^2}{4}

Area of the square is given by;

A_s = L^2

relationship between the edge length of the square, d, and length of its side, L,

d = \sqrt{L^2 + L^2} \\\\d = \sqrt{2L^2}

But area of the square , A_s = L^2

d = \sqrt{2A_s}

Then, the area of the square in terms of the edge length is given by;

A_s = \frac{d^2}{2}

Area of the circle in terms of area of the square is given by;

A_c = \frac{\pi d^2}{4} = \frac{\pi}{2}(\frac{d^2}{2} )\\\\But \ A_s = \frac{d^2}{2} \\\\A_c =  \frac{\pi}{2}(\frac{d^2}{2} )\\\\A_c =  \frac{\pi}{2}(A_s )

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Ф = E.A

Flux through the surface of the circle is given by;

\phi _{circle} = E.(\frac{\pi d^2}{4})

Flux through the surface of the square is given by;

\phi _{square} = E.(\frac{d^2}{2} )\\\\\phi _{square} =E.(\frac{d^2}{2} ).(\frac{\pi}{2} ).(\frac{2}{\pi} )\\\\\phi _{square} =E.(\frac{\pi d^2}{4} ).(\frac{2}{\pi} )\\\\\phi _{square} =(\phi _{circle}).(\frac{2}{\pi} )

Therefore, Φsquare is (\frac{2}{\pi} ) ϕcircle

5 0
4 years ago
2x + 3y is less than or equal to 15
Sholpan [36]

Answer:

less

Step-by-step explanation

sorry if I'm wrong

5 0
3 years ago
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