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1 year ago

Determine the volume of each rectangular or rectangle prism round to the nearest tenth if necessary

Mathematics

1 answer:

Svetach [21]1 year ago

3 0

The volume of the triangular prism is given by:

$\begin{gathered} V=\frac{1}{2}b\cdot h\cdot l \\ \text{Where:} \\ b=\text{base}=7.2 \\ h=\text{height}=9 \\ l=\text{length}=3 \\ V=\frac{1}{2}\cdot7.2\cdot9\cdot3 \\ V=97.2m^3 \end{gathered}$

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If f(x) = 2x + 3 and g(x) = x^2 + 1,find f(g(3)). A.) 23 B.) 47 C.) 82 D.) 90

marta [7]

The answer is A). 23.
g(x)= 2x + 3 
g(x) = x^2 + 1 
g(3) = 10 
f(g(3)) 
= f(10) 
= 23

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:

Optimizing With Lagrange Multipliers

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

How do I do this Cp geometry

ryzh [129]

Whats your geometry problem? i dont understand the question

5 0

3 years ago

Sorry for the little blur but number 18 please! And sorry for the work mess. Start where it says *Start Here*

motikmotik

You are almost done!
well, the euation of a circle centered at (h,k) and radius r is
(x-h)²+(y-k)²=r²

so you havve
(x-3)²+(y+5)²=9 (which is correct)
this can be rewritten as
(x-3)²+(y-(-5))²=3²
the center is (3,-5) and the radius is 3

5 0

3 years ago

1. Jamie purchases gifts for three friends. She chooses a book for $12.80 and then finds two identical necklaces. Her budget is

Evgen [1.6K]

Step-by-step explanation:

(48 - 12.80)/2 = $17.6

250/6 ≈ 42 and more

485 - 319 = 166 more people

14.5x + 80 > 500
14.5x > 420
x > 420/14.5
x > 29

1.25x + 3 ≤ 28
1.25x ≤ 25
x ≤ 25/1.25
x ≤ 20 miles

6. This seems a typo... 460 is likely $60

60x + 145 ≤ 625
60x ≤ 625 - 145
60x ≤ 480
x ≤ 480/60
x ≤ 8 days

7 0

3 years ago