All categories

1 year ago

What is math forlike what do we use it for whats the point of math no one uses it

1 answer:

6 0

Because,mathematics provides an effective way of building mental discipline and encourages logical reasoning and mental rigor. In addition, mathematical knowledge plays a crucial role in understanding the contents of other school subjects such as science, social studies, and even music and art.

You might be interested in

A jet plane has a takeoff speed of vto = 75 m/s and can move along the runway at an average acceleration of 1.3 m/s2. If the len

pishuonlain [190]

Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m

time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s

the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s

Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.

6 0

3 years ago

Read 2 more answers

How many 3/4 cup servings are in 12 cups of cherries?

AveGali [126]

Answer:

16 serving.

Step-by-step explanation:

Given: 3/4 cup servings.

Now, computing the number of serving in 12 cups of cherries.

Using unitary method to solve.

∴ Number of serving= $total\ cups \times number\ of\ serving\ per\ cup$

Number of serving in one cup is $\frac{4}{3}$

Number of serving in 12 cups = $12\times \frac{4}{3} = 16 \ serving$

Number of serving in 12 cups= $16\ serving$

∴ There are 16 serving is possible in 12 cups of cherries.

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

Suppose that y varies inversely with x. Use the information to find k, and then choose the equation of variation. x = 2 when y =

Sergeeva-Olga [200]

Answer:

y = $\frac{22}{x}$

Step-by-step explanation:

given that y varies inversely with x then the equation relating them is

y = $\frac{k}{x}$ ← k is the constant of variation

To find k use the condition x = 2 when y = 11

k = yx = 11 × 2 = 22

y = $\frac{22}{x}$ ← equation of variation

4 0

3 years ago

Read 2 more answers

Solve the inequality 13 > 3x + 1

Ivenika [448]

4 > x

step by step explanation:

13 > 3x + 1
-1 -1
12 > 3x
/3 /3
4 > x

3 0

2 years ago

Read 2 more answers