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Mariana [72]
3 years ago
12

How would i solve this problem ?

Mathematics
1 answer:
telo118 [61]3 years ago
5 0
You would subtract it

if the budget is lower than the amount spent, make sure to put a negative sign
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For the function <br><br> z=9x^3-4y^2+2xy, <br><br> Find dz/dx.
IceJOKER [234]

Answer:

The value of dz/dx is 27x^2 + 2y

Step-by-step explanation:

We have to given the function z = 9x^3 - 4y^2 + 2xy and we have to find the partial derivative of the given function. In the given function, we have to find partial derivation with respect to x. Therefore, the other values will be considered as constant and during the partial derivation, the constant value will become zero.

z=9x^3-4y^2+2xy \\\frac{dz}{dx} = 9\times 3x^{3-1} -0 +2y \\\frac{dz}{dx} =27x^2 + 2y \\

Therefore, the value of dz/dx is 27x^2 + 2y

4 0
4 years ago
If u 500 tens 60ones how many do u have in all???
Tomtit [17]
If this is in value, you would have $5,060. If you mean bills it would be 560.
3 0
4 years ago
Please help with this question please
liberstina [14]

Answer:

31. 20

Step-by-step explanation:

Each class had 40 students right?...so

4 caring=40-36

<h3>4caring=4 students</h3>

<h3>4Dutiful=40-40</h3>

=0 (no student had absent for this class)

4excellence=40-29

<h3>=11 students</h3>

4 friendly=40-35

<h3>=5 students</h3>

Now plus all the number together:

4+11+15=

Final answer:20 students

7 0
3 years ago
Is it possible for a composite number to have more than one prime factorization? Yes. I'm I right @dirtydan667
-Dominant- [34]
<span>Is it possible for a composite number to have more than one prime factorization?The Answer is Yes. Prime factors are factors of a composite number that are indivisible except by the number 1 or the number itself. it is possible especially for very large numbers. 2&3. No, because as mentioned previously, the default prime factors of numbers are 1</span>
3 0
3 years ago
A school director must randomly select 6 teachers to participate in a training session. There are 34 teachers at the school. In
anastassius [24]
The formula for the number of combinations when choosing "r" objects (or people) from a larger set of "n" objects (or people) is:
combinations = n! / [r! * (n-r)!]
combinations = 34! / [6! * 28!]
combinations = 34*33*32*31*30*29*28! / [6! * 28!]
combinations = 34*33*32*31*30*29 / 6*5*4*3*2
combinations = 34*33*32*31*29 / 4*3*2
combinations = 34*11*4*31*29
combinations = 1,344,904

Source:
http://www.1728.org/combinat.htm



3 0
3 years ago
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