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2 years ago

Help help help help help

Mathematics

2 answers:

attashe74 [19]2 years ago

7 0

Answer:

i dont now

Step-by-step explanation:

inde man kona ka balo sorry

Dafna1 [17]2 years ago

3 0

Answer:

quadrant 1 and quadrant IV

Step-by-step explanation:

(a)

secθ > 0 in quadrants 1 and IV

sinθ > 0 in quadrants 1 and II

Thus θ is in quadrant I

(b)

cosθ > 0 in quadrants I and IV

cotθ < 0 in quadrants II and IV

Thus θ is in quadrant IV

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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

Greatest common divisor of 60 and 792 using the prime factorization method

klio [65]

Answer:

Step-by-step explanation:

8 0

3 years ago

Please help!

AfilCa [17]

Answer:

a.) P(HH|C)=2/5

Explanation:

Given following:

P(HH) = 7

P(C) = 10

P(HH ∩ C) = 4

P(HH ∪ C) = 13

Solving steps:

$\sf P(HH|C) = \dfrac{P(HH \cap C) }{P(C)}$

$\sf P(HH|C) = \dfrac{4 }{10}$

$\sf P(HH|C) = \dfrac{2 }{5}$

8 0

1 year ago

Select the correct answer from each drop down menu. Consider the given table.

svet-max [94.6K]

Answer:

the new year will also......so the equation of a new 6 month period will not have to go on coronavirus or the equation is 4 year old but it was a very difficult task for me and my students and the other side of it to do it and I was the first to do it in your own way to get a better job in your life will I am a good thing to be a teacher in a good thing and the other side of it you will have the opportunity to be the best in your career in the world and in your life will you have been

5 0

3 years ago

A greenhouse is offering a sale on tulip bulbs because they have inadvertently mixed pink bulbs with red bulbs. If 40% of the b

True [87]

Answer:

The probabilty of having at least one pink bulb if 4 bulbs are purchased is 13%.

Step-by-step explanation:

The proportion of pink bulbs is p=0.4.

If X is the amount of pink bulbs in a group of 4 bulbs, we can model this as a binomial distribution problem.

The probability that at least one of the bulbs is pink is P(X≥1)

$P(X\geq 1)=1-P(X=0)=1-(\binom{4}{0}p^0(1-p)^4)\\\\P(X\geq 1)=(1-p)^4=0.6^4=0.1296$

The probabilty of having at least one pink bulb if 4 bulbs are purchased is 13%.

6 0

2 years ago