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3 years ago

Helppp meeee plssss with this pls

Mathematics

1 answer:

andre [41]3 years ago

6 0

Your answer is 10.65

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last year the cost of a season ticket for a rugby club £370. This year the cost of a season ticket for the club has been increas

Elden [556K]

Answer:

$\frac{80}{370}$ OR $\frac{8}{37}$

Step-by-step explanation:

So first you find the difference between the new cost and the old cost:

450 - 370 = 80

So the increase was 80.

When you write it as a fraction you write the increase ( 80 ) over the starting cost ( 370 ).

Thus the answer becomes $\frac{80}{370}$

But you can always reduce it to its lowest terms thus:

$\frac{8}{37}$

HOPE THIS HELPED

4 0

3 years ago

List intercepts for 4x^2+9y^2=36

polet [3.4K]

X intercept = (-3, 0) (3, 0)
y int = (-2,0) (2,0)

5 0

3 years ago

Read 2 more answers

How many faces does the solid made from the net below have?<br>

nikdorinn [45]

Answer:We need to see the net.

Step-by-step explanation:

5 0

3 years ago

An ac unit uses 3500 watts or power. If it runs for 10 hours each day and the utility company charges $0.13 per kWh, how much do

Helga [31]

Cost = (3500 W)·(1 kw/1000 W)·(10 hrs/day)·(30 day/month)·($0.13/kw-hr)

Cost = (3500·1·10·30·0.13 / 1000)·(W·kW·hr·da·$ / W·da·month·kw·hr)

Cost = <em> 136.50 $/month</em>

5 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).