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2 years ago

HELLOOOOOooOOOOOooOOoooOOoo

Mathematics

1 answer:

Viefleur [7K]2 years ago

6 0

Answer:

6. Multiply 2 by 6

7. 71 + 9(8) = 71 + 72 = 143

8. 52 + 3(4) = 12 + 52 = 64

9. 19.25 + 2.75(7) = 19.25 + 19.25 = 38.50

10. 2(0.6)^2 + 4(0.6)(5) = 0.72 + 12 = 12.72

Srry if they r wrong

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Please help meeeeeeeee

Olegator [25]

Answer:

Step-by-step explanation:

4 0

2 years ago

Can someone help me please !!!!!!!!!!!!!!!!!!

Reptile [31]

Answer:

y = -1x +6

Step-by-step explanation:

6 0

2 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

Which set of ordered pairs has point symmetry with respect to the origin (0, 0)?

Leya [2.2K]

Answer:

(-8, 3), (8, -3)

Step-by-step explanation:

Point symmetry about origin means reflection of the given point about the origin.

The reflection of a point about the origin will cause the 'x' and 'y' value of the point to change its sign.

Therefore, the coordinate rule for point symmetry about the origin is given as:

$(x,y)\to (-x,-y)$

Now, let us check each of the given options.

Option 1:

(-8, 3), (8, -3)

Now, if (x, y) = (-8, 3), then its point symmetry is given as (-(-8), -3) = (8, -3)

So, option 1 is correct.

Option 2:

(-8, 3), (-3, 8)

Now, if (x, y) = (-8, 3), then its point symmetry is given as (-(-8), -3) = (8, -3) ≠ (-3, 8)

So, option 2 is not correct.

Option 3:

(-8, 3), (-8, -3)

Now, if (x, y) = (-8, 3), then its point symmetry is given as (-(-8), -3) = (8, -3) ≠ (-8, -3)

So, option 3 is not correct.

Option 4:

(-8, 3), (8, 3)

Now, if (x, y) = (-8, 3), then its point symmetry is given as (-(-8), -3) = (8, -3) ≠ (8, 3)

So, option 4 is not correct.

Hence, only option 1 is correct.

5 0

3 years ago

Mary had $156.00 and spent 1/6 of it for a coat. She then bought a skirt with 1/4 the money she had left. How much money did she

marusya05 [52]

Answer: $ 97.5

Step-by-step explanation:

156 * 1/6 = 26

156-26=130

130 * 1/4- 32.5

130-32.4= 97.5

3 0

2 years ago