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2 years ago

Select ALL the correct answers.

1 answer:

7 0

Expression A decays at a rate of 5% every three months, while Expression B grows at a rate of 12% every four months and Expression A has an initial value of 624, while expression B has an initial value of 725 , Option B and E are the correct statements.

<h3>What is an Expression ?</h3>

It is a mathematical statement tat consists of variables , constants and mathematical operators.

It is given that

The company's records show the sales of short boards decrease every three months as represented by expression A, where t is the number of years that the boards have been for sale.

The company's records show that the sales of long boards increase every four months as represented by expression B, where t is the number of years that the boards have been for sale.

$\rm Expression\; A = 624(0.95)^{4t}\\\\Expression \; B = 725(1.12)^{3t}$

From observing both the expression the initial value is when t = 0

when t = 0,

A = 624

B = 725

and so Expression A decays at a rate of 5% every three months, while expression B grows at a rate of 12% every four months and Expression A has an initial value of 624, while expression B has an initial value of 725 , Option B and E are the correct statements.

To know more about Expression

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Eugene wants to buy jeans at a store that is giving $10 off everything the tag on the gene is marked 50% off the original price

Blizzard [7]

Hello!

Subtract the $10 discount first:

49.98 - 10 = 39.98

Next, divide this price in 2 because 50% means half off:

39.98 / 2 = 19.99

So, the total cost of the jeans, after 2 discounts have been applied, is $19.99.

7 0

3 years ago

Bruce owns a business that produces widgets. He must bring in more in revenue than he pays out in costs in order to turn a profi

Aliun [14]

I believe the answer is 267

If each is worth $25 and it costs $10 then he is making $15 profit.

4000/15 is 266.6666667

So you would need 267

8 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).