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inna [77]
3 years ago
14

You can compare irrational numbers using rational approximation, yes or no?

Mathematics
2 answers:
lord [1]3 years ago
6 0

Answer:

yes

Step-by-step explanation:

loris [4]3 years ago
4 0

Answer:

YES!!

Step-by-step explanation:

You might be interested in
X
ratelena [41]
C the correct answer is C
4 0
3 years ago
What is the length of line segment KJ?<br><br> 2√3 units<br> 3√2units<br> 3√3units<br> 3√5units
Pavlova-9 [17]
KJ^2=6^2+3^2
KJ=the square root of 36+9
KJ=square root of 45
kJ= 3 times the square root of 5
The answer is D
3 0
3 years ago
Read 2 more answers
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
Kyra has 2 plates, 2 cups, and 2 bowls. If she chooses one of each randomly, what is the probability that the plate, cup, and bo
tensa zangetsu [6.8K]
The correct answer is A. <span>0.125
</span><span>
1/8(this is a fraction)</span>
(divide the numerator by the denominator)

1÷8 = 0.125
If you want you can Convert the number to a percentage:<span>
0.125×100=12.5%

but the question is asking for decimal form therefore the answer is </span>0.125.
I also did the quiz and this was correct

Hope this helps!! :D
7 0
3 years ago
Read 2 more answers
If m FDG = 50°, what is m FEG?<br><br> 75°<br> 50°<br> 100°<br> 25°<br><br> Thank you!
denpristay [2]
The measure of the angle FEG is probably the same as the measure of the angle FDG. There is nothing saying that EF is parallel to DG, but if so, the measure of the angle FEG is also 50º.
When the diagonals of a trapezium with two parallel bases inside of a circle are drawn, they make the same angle measure with those bases.
7 0
3 years ago
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