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ivolga24 [154]
3 years ago
6

Simplify without negative exponents. x^-4 x^7 =is It x^11?

Mathematics
1 answer:
Gwar [14]3 years ago
7 0

We have x^(-4) • x^(7).

This leads to x^(-4 + 7).

Work out the math inside the parentheses.

ANSWER: x^(3).

Did you follow?

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What is the solution to the following system
Archy [21]

C is the answer.............

3 0
3 years ago
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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).

Here's a comprehensive list of all the critical points we found:

(\sqrt[4]{13},0,0)

(-\sqrt[4]{13},0,0)

(0,\sqrt[4]{13},0)

(0,-\sqrt[4]{13},0)

(0,0,\sqrt[4]{13})

(0,0,-\sqrt[4]{13})

\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

5 0
3 years ago
Help me figure this out
Semenov [28]

Answer:

Each number matches up below. The left side is x, and the right is y.

1 ⇒ 192

2 ⇒ 384

4 ⇒ 768

6 ⇒ 1152

10 ⇒ 1920

Step-by-step explanation:

We know in 1 serving, there is 192 calories. Following this, to find the calories for 2 servings, we multiply 192 by 2.

192*2 = 384

When we know the calories but not the serving, you divide the calories by 192

768/192 = 4

6 0
3 years ago
4.) Find the quotient<br><br> 20 ÷ 1/4<br><br><br> A.) 5<br> B.) 40<br> C.) 10<br> D.) 80
Mazyrski [523]

1/4 of 20 is 5

A is the answer

7 0
3 years ago
1/3 of $6.00 <br> What is the new price? <br> Show how you got your answer please!
Katarina [22]

Answer:

$2

Step-by-step explanation:

1/3 of $6 = 1/3 * 6 = $2.00

5 0
3 years ago
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