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Sauron [17]
3 years ago
11

2 + 5.8 = 6 why is this incorrect

Mathematics
2 answers:
Marat540 [252]3 years ago
7 0

Answer:

6

Step-by-step explanation:

Because you add 5.8

                        <u>+        2 </u>

                              6.0

you add 8+2=10 you put the zero and then you add the 1 to the 5 and is = 6

Tanzania [10]3 years ago
3 0

This is incorrect because 2 + 5.8 = 7.8. And 7.8 does not equal 6. 2 + 4 = 6 is the correct answer. This could be represented by this:

2 + x = 6.

Subtract two from both sides to get:

x = 4.

When substituted back into the equation:

2 + (4) = 6.

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Simplify the expression <br> 3(5x+8)
viva [34]

Answer:

Solution given:

3(5x+8)

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15x+8*3

<u>1</u><u>5</u><u>x</u><u>+</u><u>2</u><u>4</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u><u>.</u>

3 0
3 years ago
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Box 1 contains 1000 lightbulbs of which 10% are defective. Box 2 contains 2000 lightbulbs of which 5% are defective. (a) Suppose
mojhsa [17]

Answer:

a) There is a 66.7% chance that you were given box 1

b) There is a 80% chance that you were given box 1

Step-by-step explanation:

To find this, we need to note that there is a 1/10 chance of getting a defective bulb with box 1 and a 1/20 chance in box 2.

a) To find the answer to this, find the probability of getting a defective bulb for each box. Since there is only one bulb pulled in this example, we just use the base numbers given.

Box 1 = 1/10

Box 2 = 1/2

From this we can see that Box 1 is twice as likely that you get a defective bulb. As a result, the percentage chance would be 2/3 or 66.7%

b) For this answer, we need to square each of the probabilities in order to get the probability of getting a defective one twice.

Box 1 = 1/10^2 = 1/100

Box 2 = 1/20^2 = 1/400

As a result, Box 1 is four times more likely. This means that it would be a 4/5 chance and have a probability of 80%

3 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
3 years ago
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xxTIMURxx [149]

Answer:

1. MIGHT be 16 1/2

Step-by-step explanation:

causse 2/3 / 4/9 = 1 1/2

1 1/2 + 10= 11 +5 = 16

16+1/2 = 16 1/2

4 0
2 years ago
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(3m^3- 1/2 y) (3m^w - 1/2y)<br> multiply
Westkost [7]

Answer:

This is just multiplying not simplify.

4 0
2 years ago
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