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olganol [36]
3 years ago
9

A car travels at a constant speed of 40 mph on the highway.

Mathematics
1 answer:
Alexus [3.1K]3 years ago
8 0

Answer:no

Step-by-step explanation:the trafic may slow down or stop

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PLEASE HELP I WILL MARK BRAINLIEST explanation please!! :)
julia-pushkina [17]

Answer: c

Step-by-step explanation:

u want to separate it into the two triangles and a rectangle. so the first marked triangle is the first part of the problem then it adds on the rectangle 6(12) then theres that other unmarked triangle because u can see the bottom is 14 when the top is 12 so u take the formula and fill in 1/2(6)(2). i hope this makes sense im bad at explaining things.

5 0
3 years ago
Read 2 more answers
5n(3n-n+8) slove it please
lozanna [386]
Hey there! :) 

5n(3n - n + 8)

Simplify.

(5n × 3n) + (5n × -n) + (5n × 8)

Simplify.

(15n²) + (-5n²) + (40n)

Remove parenthesis

**Remember : a negative times a positive is always a negative, a positive times a positive is a positive, and a negative times a negative is a positive**

15n² - 5n² + 40n

Combine like terms.

(15n² - 5n²) + 40n

Simplify.

10n² + 40n → final answer :) 

~Hope I helped!~
4 0
3 years ago
The gas co2 is diffusin at steady state through a tube 20 cm long having a diameter of 1 cm and containing n2 at 350 k . the tot
Marrrta [24]

The value is J = 1.71*10^-6 kmol/m^2.s

According to the given conditions we get,

The length the tube l= 0.20m

The diameter of the tube d= 0.01m

The total pressure inside the tube P= 101.32kPa

The partial pressure of CO2 at the first end is

P1= 456mm Hg = 60794.832 Pa

The partial pressure of CO2 at the other end is

P2= 76mm Hg = 10132.472 Pa

The temperature is T = 298 K

The diffusion coefficient D= 1.67* 10^-5 m^2/s

Generally the molar flux of CO2 is mathematically represented as

J= D{p1-p2} / R.T

Here R is the gas constant with value R= 8.314 j/kmol

So we get J= 1.71 * 10^-3 mol/m^2.sec

Learn more about Mole on:

brainly.com/question/27270616

#SPJ4

       

       

4 0
2 years ago
What is the expand form for 0.632
pav-90 [236]
0.600 is ur answer Hopes this helps u.
5 0
3 years ago
Locate the foci of ellipse. Show your work.<br> X^2/36 +y^2/11 =1
Eva8 [605]
<span>In this equation, 36>11 and so the ellipse has major axis along x axis
standard equation of ellipse with major axis is along x axis
x^2/a^2+y^2/b^2=1,
the foci are at (ae,0) and (–ae,0)
eccentricity e is given by
ae=√(a^2-b^2)

In this case ae=√(36-11)= √25=+/- 5

Focii are(-5,0) and(5,0) </span>
6 0
3 years ago
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