All categories

2 years ago

What is -0.9 if I turn it into a fraction in simplist form?

Mathematics

2 answers:

Arada [10]2 years ago

7 0

-9/10. Hope that helps!

vodka [1.7K]2 years ago

7 0

0.1 is the same as 1/10. Therefore, 9 x 0.1 = 9 x 1/10 = 9/10.
Since it is negative, the answer is -9/10.

You might be interested in

What is the surface area of the triangular pyramid?

KonstantinChe [14]

Answer:

126 Sq unite

Step-by-step explanation:

I took the assignment

8 0

2 years ago

Read 2 more answers

Write 77/20 as a terminating decimal

ahrayia [7]

77/20 = 3.85 <== terminating decimal

8 0

2 years ago

Sally makes a fruit drink using ⅔ of a cup of orange juice for every ¼ of a cup of pineapple juice. How many cups of orange juic

Shkiper50 [21]

Answer:

5 1/3

Step-by-step explanation:

Set up ratio.

2/3 cup orange juice X 2 cups pineapple juice

———————————-

1/4 cup pineapple juice

first multiply numerator : 2/3 x 2 = 4/3

and divide by denominator

(4/3)/(1/4). Keep the first. Change the sign to multiplication and Flip the second fraction.

4/3 x 4/1 = 16/3

change to proper fraction : 16/3 = 5 1/3 cups

7 0

3 years ago

The formula for the surface area of a cylinder is 277(r + h) where r is the radius of the cylinder's base and h is its height. A

Andrew [12]

Answer:

277(10+2)=277(12)=277x12=3324

8 0

2 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:

$g: x_1+x_2+x_3=1000$

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

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

3 years ago