How many seconds are in 14 years?

A single gram of a certain metallic substance has 0.52 g of copper and 0.25 g of zinc the remaining portion of the substance is

andrezito [222]

Ben's estimate gives 7 g of nickel; the actual amount is 8.03 g.

In 1 g of the substance, there is 0.52 g of copper and 0.25 g of zinc; this gives
0.52+0.25 = 0.77 g of the substance.

The remaining part of the substance is nickel:
1-0.77 = 0.23 g of nickel.

Using Ben's estimate, 0.2 g of nickel per gram of substance, we have
0.2(35) = 7 g of nickel in 35 g of the substance.

The actual amount is 0.23(35) = 8.03 g of nickel in 35 g of the substance.

6 0

3 years ago

Could anyone help me out, it’s would mean a lot.

SIZIF [17.4K]

Answer:

$x = 19$

Step-by-step explanation:

If two lines are perpendicular, they create 4 90° angles. Meaning, in this case, that m<DBC = 90 and m<DBA = 90.

We're given that m<DBE = 2x - 1 and that m<CBE = 5x - 42.

The sum of angles DBE and CBE = m<DBC = 90°

We can add the two angles and set it equal to 90 to find x

$2x - 1 + 5x - 42 = 90\\7x - 43 = 90\\7x = 133\\x = \frac{133}{7} =19$

6 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

FIRST PERSON TO ANSWER THIS CORRECTLY GETS BRAINLEIST

cupoosta [38]

Answer:

Nine less than twice a number 2x-9

Seven divided by a number 7-x

Ninety-nine plus the difference of eleven and five 99+(11-5)

Four-fifths more than three times. Number 3x+4/5

Step-by-step explanation:

Please mark brainliest

5 0

3 years ago

Read 2 more answers

Given the function g(x) = 6(4)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Natasha2012 [34]

I also need this answer so when you find out let me know

5 0

3 years ago