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Mrac [35]
3 years ago
15

What do you call moving digits around to a different place when you count up or down past 9?

Mathematics
2 answers:
pickupchik [31]3 years ago
8 0
D is the answer. All the above.

Happy Cheating!! You will get a good grade
Natasha2012 [34]3 years ago
4 0
All of the above is the answer
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Can someone help me and don't comment if you don't know the answerer your just taking points from someone who dose know it
Svetach [21]
Original equation x = yz / a solve for a

xa = yz
a = yz / x
3 0
4 years ago
What is the distributive property for 9(72)
spayn [35]

9(72)= 648 will be your answer to solve this problem but 9 is your distributive property


8 0
4 years ago
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If Tina traveled 1,000 miles in 5 hours on her plane flight to her vacation destination, how fast was the plane traveling? A.200
belka [17]
OK so for this it will be simple 
1,000 divided by 200 is 5 
so if the plane took 5 hours to do 1,000 miles then the plane was traveling at option A. 200 miles per hour.
4 0
3 years ago
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Verify that the conclusion of Clairaut’s Theorem holds, that is, uxy = uyx, u=tan(2x+3y)
choli [55]

Answer: Hello mate!

Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A

for all the elements in A, the, for all the elements on A you get:

\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)

This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.

Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:

\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}

\frac{du}{dx}  =  \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}

and now lets derivate this with respect to y.

using that \frac{d(sec(x))}{dx}= sec(x)*tan(x)

\frac{du}{dxdy} = \frac{d(2*sec(2x + 3y)^{2} )}{dy}  = 2*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*3 = 12sec(2x + 3y)^{2}tan(2x + 3y)

Now if we first derivate by y, we get:

\frac{du}{dy}  =  \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}

and now we derivate by x:

\frac{du}{dydx} = \frac{d(3*sec(2x + 3y)^{2} )}{dy}  = 3*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*2 = 12sec(2x + 3y)^{2}tan(2x + 3y)

the mixed partial derivates are equal :)

7 0
3 years ago
The Celsius temperature is 32 degrees. What is the Fahrenheit temperature
Aleksandr [31]

Answer:

89.6

Step-by-step explanation:

(32°C × 9/5) + 32 = 89.6°F

3 0
3 years ago
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