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Leviafan [203]
3 years ago
14

Will mark brainliest Please help with these two questions for me

Mathematics
2 answers:
USPshnik [31]3 years ago
7 0

Answer:

Go ahead give him Brainlyist I need it but he can have since he is really good

Step-by-step explanation:

My name is Ann [436]3 years ago
4 0

refer to attachments

  • Attachment 1=Solution 1
  • Attachment 2=Solution 2

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Which mixed number is equivalent to the improper fraction below?
Solnce55 [7]

D is the correct fraction

5 0
3 years ago
Solve the following linear system by graphing <br> 4x+5y=10<br> 3x-3y=21
Inga [223]

Answer:

y = 2

x = 5

Step-by-step explanation:

4x + 5y = 10 | ×3 |

3x - 3y = 21 | ×4 |

12x + 15y = 30

12x - 12y = 84

____________--

27y = -54

y = -54/27

y = 2

4x + 5y = 10 | ×3 |

3x - 3y = 21 | ×5 |

12x + 15y = 30

15x - 15y = 105

____________+

27x = 135

x = 135/27

x = 5

6 0
3 years ago
Convert 132 pounds to kilograms. Use 1 kg = 2.2 lb.
Vera_Pavlovna [14]

1kg = 2.2lb

1lb = 0.454kg

132lb = 0.454*132

132lb = approx 60kg

3 0
3 years ago
What is the collision of the line that passes through the points (-2,3) and (2,7)
Nina [5.8K]
I don't know, I just need points 

lol
8 0
4 years ago
It is desired to draw blood 3 cm up a capillary tube. Assuming a contact angle of 0 degrees and a blood density of 1060 kg/m3, w
JulijaS [17]

Answer:

If we assume a temperature of 20ºc and the blood interfacial surface tension is similar to water interfacial surface tension, the diameter of the capillary tube should be 0.933mm.

Step-by-step explanation:

The Jurin law describes the height a fluid can reach in a capillary tube. This law can be written as:

H=\displaystyle\frac{2\gamma cos(\theta)}{\rho gr}

where γ is the interfacial surface tension, θ is the contact angle with the fluid, ρ is the fluid density, g is the gravity acceleration and r is the tube radius.

If we assume that the interfacial surface tension of blood and water are almost the same, γ=0,0728 N/m at 20ºc. Therefore the diameter of the tube will be:

D=2r=\displaystyle\frac{4\gamma cos(\theta)}{\rho gH}=0.933\cdot 10^{-3}m

4 0
3 years ago
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