All categories

3 years ago

Which statement about rational numbers is true?

Mathematics

2 answers:

Alina [70]3 years ago

8 0

Answer:

D.All fractions containing integers are rational numbers

Step-by-step explanation:

Misha Larkins [42]3 years ago

7 0

Answer:

D: All fractions containing integers are rational numbers.

You might be interested in

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

Evaluate the expression when x = 3.<br><br> x² + 6x - 3

Black_prince [1.1K]

Answer:

X=3

${3}^{2} + 6(3) - 3 \\ 9 + 18 - 3 \\ 27 - 3 \\ = 24$

3 0

2 years ago

Rewrite the expression 15(8 - 2) using the distributive property.

Charra [1.4K]

Answer:

Step-by-step explanation:

15(8-2)

15-(6)

4 0

3 years ago

Read 2 more answers

15 players two coaches and 21 parents attend this year sports banquet the cost of each meal was $12 +5% sales tax what was the t

Artist 52 [7]

Answer:

$478.80

Step-by-step explanation:

We have been given that 15 players two coaches and 21 parents attend this year sports banquet the cost of each meal was $12 plus 5% sales tax.

First of all, we will find cost of each meal with tax as:

$\text{Cost of each meal with tax}=\$12+\$12\times \frac{5}{100}$

$\text{Cost of each meal with tax}=\$12+\$12\times 0.05$

$\text{Cost of each meal with tax}=\$12+\$0.6$

$\text{Cost of each meal with tax}=\$12.6$

Now, we will find total persons at the sport banquet as:

$\text{Total persons at banquet}=15+2+21$

$\text{Total persons at banquet}=38$

To find the total cost of food for the banquet, we will multiply cost of food for each person by 38.

$\text{Total cost of food for the banquet}=\$12.6\times 38$

$\text{Total cost of food for the banquet}=\$478.8$

Therefore, the total cost of food for the banquet is $478.80.

5 0

3 years ago

What is 2x-{2-[2(x-2)-(2-x)]}=0

Ludmilka [50]

$2x-(2-(2(x-2)-(2-x)))=0\\
2x-(2-(2x-4-2+x))=0\\
2x-(2-(3x-6))=0\\2x-(2-3x+6)=0\\
2x-(-3x+8)=0\\
2x+3x-8=0\\
5x=8\\
x=\dfrac{8}{5}$

3 0

3 years ago

Which statement about rational numbers is true?​

Which statement about rational numbers is true?