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3 years ago

HELP PLS THIS IS HARD

Mathematics

1 answer:

BaLLatris [955]3 years ago

5 0

Answer:

its the second one

Step-by-step explanation:

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Nutka1998 [239]

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3 years ago

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pav-90 [236]

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3 years ago

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A garden supply store sells two types of lawn mowers. Total ales of mowers for the year were $8379.70. The total number of mower

Anestetic [448]

Answer:

The number of small mowers are 19 and the large mowers are 11.

Step-by-step explanation:

Given:

A garden supply store sells two types of lawn mowers. Total sales of mowers for the year were $8379.70. The total number of mowers sold the 30. The small mower cost $249.99 and the large mower costs $329.99.

Now, to find the number of each type of mower sold.

Let the number of small mower be $x.$

And the number of large mower be $y.$

So, total number of mowers are:

$x+y=30$

$x=30-y\ \ \ ....(1)$

Now, the total sales of mowers are:

$249.99(x)+329.99(y)=8379.70$

Substituting the value of $x$ from equation (1) we get:

$249.99(30-y)+329.99y=8379.70$

$7499.7-249.99y+329.99y=8379.70$

$7499.7+80y=8379.70$

<em>Subtracting both sides by 7499.7 we get:</em>

$80y=880$

<em>Dividing both sides by 80 we get:</em>

$y=11.$

The number of large mower = 11.

Now, to get the number of small mowers we substitute the value of $y$ in equation ( 1 ):

$x=30-y\\x=30-11\\x=19.$

The number of small mower = 19.

Therefore, the number of small mowers are 19 and the large mowers are 11.

6 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).