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bazaltina [42]
2 years ago
8

1 to the 999999999999 power

Mathematics
1 answer:
Reil [10]2 years ago
4 0

Answer:

1

Step-by-step explanation:

1 times anything equals the opisite number but in this case you are multiplying 1 by itself 999999999999 times and that equals 1

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PLZ HELP I WILL GIVE BRIANLY!
kumpel [21]

Answer:36

Step-by-step explanation:

\frac{4}{7} =\frac{48\\}{84}

for every 7 students there are 4 girls, so for every 84 there must be 48

7*12=84

4*12=48

so, in all 84 students there are 48 girls

so amount of boys=84-48=36

there are 36 boys

8 0
3 years ago
12/4 = 12 1/4 True or False
Stells [14]

Answer:

False

Step-by-step explanation:

12/4=3

3 does not equal 12 1/4

6 0
3 years ago
Between which pair of numbers is the exact product of 379 and 8?
AnnZ [28]
The product of 379 and 8 is 3,032.

It's between (any number less than 3,032) and (any number greater than 3,032).

I guess if this is an <em>estimation </em>exercise, you could say that it's between
(8 x 300) and (8 x 400), or 2,400 and 3,200. 
5 0
3 years ago
Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o
Ivahew [28]

Answer:

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

\displaystyle C_i=3x_i+\frac{i}{40}x_i^2

It means the cost for each generator is expanded as

\displaystyle C_1=3x_1+\frac{1}{40}x_1^2

\displaystyle C_2=3x_2+\frac{2}{40}x_2^2

\displaystyle C_3=3x_3+\frac{3}{40}x_3^2

The total cost of production is

\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2

Simplifying and rearranging, we have the objective function to minimize:

\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)

The restriction can be modeled as a function g(x)=0:

g: x_1+x_2+x_3=1000

Or

g(x_1,x_2,x_3)= x_1+x_2+x_3-1000

We now construct the auxiliary function

f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)

\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)

We find all the partial derivatives of f and equate them to 0

\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0

\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0

\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0

f_\lambda=x_1+x_2+x_3-1000=0

Solving for \lambda in the three first equations, we have

\displaystyle \lambda=3+\frac{2}{40}x_1

\displaystyle \lambda=3+\frac{4}{40}x_2

\displaystyle \lambda=3+\frac{6}{40}x_3

Equating them, we find:

x_1=3x_3

\displaystyle x_2=\frac{3}{2}x_3

Replacing into the restriction (or the fourth derivative)

x_1+x_2+x_3-1000=0

\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0

\displaystyle \frac{11}{2}x_3=1000

x_3=181.8\ MW

And also

x_1=545.5\ MW

x_2=272.7\ MW

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

5 0
4 years ago
How many 1/2 kg packets of salt can you obtain from 18 kg of salt​
True [87]

Answer:

36

Step-by-step explanation:

18 × 2 = 36

Because 1/2 is half of 1 kg so multiply by 2 it makes 1 kg

5 0
2 years ago
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