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3 years ago

Round to the nearest hundredth

Mathematics

1 answer:

Snezhnost [94]3 years ago

8 0

Answer: 0.14

Step-by-step explanation:

5 / (25+5+5)

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What is the equation of the line y-intercept is 3 and the slope is 1

Hoochie [10]

Answer:

y=x+3

Step-by-step explanation:

y=mx+b

where m is the slope and b is the y-intercept so

y=1x+3

simplified

y=x+3

6 0

3 years ago

What is the volume of a cone that has a height of 7 mm and a radius of 3 mm? Use 3.14 to approximate pi and round the answer to

Anna [14]

Volume=1/3 times area of base (circle) times height

circle=3.14 times radius^2
circle=3^2 times 3.14
circle=9 times 3.14=28.26
times height (7)
197.82
times 1/3=65.94
answer is 65.94 mm^3

3 0

3 years ago

How do you write the decimal expansion for 6/11

xxMikexx [17]

Divide 6 by 11

6 / 11 = 0.5454545455

Answer is A

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3 years ago

Read 2 more answers

How many terms are in the algebraic expression<br><br>Also, What do they mean by "Terms"

kari74 [83]

Answer:

There are 4 terms

Step-by-step explanation:

A term is a single mathematical expression. Terms can be identified with a plus or minus sign in front of them. Terms can also be multiplied and divided.

So, in this case, the terms are:

-7

12x^4

-5y^8

4 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).