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3 years ago

Which equations have the same value of x as Two-thirds (6 x + 12) = negative 24? Select two options

Mathematics

2 answers:

Valentin [98]3 years ago

7 0

Answer:

-36

Step-by-step explanation:

Brainliest OR I WILL FIND YU

OLga [1]3 years ago

6 0

Answer:

-6

Step-by-step explanation:

so all you have to do is subtract the 12 from the -24. to get -36.

leaving you with this:

6x = -36

then all you do is divide the 6x from the -36, to get -6.

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A patient is ordered 1.6 g Drug A every 8 hours over a 24

Vlad [161]

Answer:

33.6 grams

Step-by-step explanation:

Rate 1;

8 hours = 1.6 g

24 hours = 24 × 1.6/8 = 4.8 hours

Rate 2;

4 hours = 1.6 g

72 hours = 72 × 1.6/4 = 28.8 g

Total grams of Drug A that the patient receive

altogether = 28.8 + 4.8 = 33.6 g

5 0

2 years ago

What is the value of the expression 3² .(2³+4)/2²

Ber [7]

Answer:

Step-by-step explanation:

3^2(2^3+4)/2^2

3^2(8+4)/2^2

3^2(12)/2^2

9(12)/4

108/4

4 0

3 years ago

If 2 sides of a triangle measure 6 and 16 find the range for the 3rd side

tangare [24]

Step-by-step explanation:

Range if third side is in between 10-22

Reason:

For the lowest value of third side 16-6= 10

For the highest value if third side 16+6= 22

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8 0

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

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).