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3 years ago

Explain why -11 is less than -7, but |-11| is greater than |-7|

Mathematics

2 answers:

kolezko [41]3 years ago

6 0

Because the absolute value makes anything positive so -11 to 11 and -7 to 7 which would mean that 11 >7 and -11<-7

Ber [7]3 years ago

5 0

When looking at the -7 and -11 it's like looking at a timeline you see the numbers from left to right so you would look at the -11 first and then the -7. whenever a number has the two lines it's the absolute value so you would ignore the negative signs and it would just be the number.

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Plz help 52 points

Troyanec [42]

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3 years ago

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Salsk061 [2.6K]

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3 years ago

compute the exact value of the function for the given x-value wothout using a calculaotor . f(x)=(1/3)^x for x=-1

ANEK [815]

Answer:

f(x) = (1/3)^x

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= 3

Hope this helps!

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3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

$\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0$

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

4 0

3 years ago

A tournament is being run between two teams, A and B. This is a 2- tournament meaning that the first team to win 2 games is the

natta225 [31]

Answer:

Step-by-step explanation:

For first game PA = 1/3

For second game PA = 1/3 ( If A is not lost in first game )

= 2/5 (If A is lost in first game )

For conclusion of game in three matches :

If A wins , the probability is

PA , PB , PA = 1/3 X 2/3 X 3/5 = 6/45

PB , PA , PA = 2/3 X 3/5 X 1/3 = 6/45

If B wins

PA, PB, PB = 1/3 X 2/3 X 2/5 = 4/ 45

PB, PA , PB = 2/3 X 3/5 X 2/3 = 4 / 15

Total probability of conclusion of game in 3 matches

= 6/45 +6/45 + 4/45 +4/15 = 28/45

b )

For the game concluding in 2 matches , the probability are as follows

PA,PA = 1/3 X 1/3 = 1/9

PBPB = 2/3 X 2/5 = 4 / 15

Total probability

= 1/9 + 4/15 = 17/45

So PA = 1/9 / 17/45

= 5/17

5 0

3 years ago