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3 years ago

Kindly Please Help....

Mathematics

2 answers:

Nuetrik [128]3 years ago

7 0

Sam is a little fish. He lives in the sea. He is very lonely. He wants to have a friend. The friend looks like him. Sam sees an ink fish. The ink fish has eight legs. He doesn't look like Sam. So Sam goes away. Sam meets a shark. He wants to say hello to the shark. The shark opens his big mouth. Sam runs away quickly. Sam is tired and hungry. He wants to have a rest. Then he sees a round fish. She says to him. "Hello! Would you like to be my friend?" Sam answers: "Of course! But you are round. I am flat." The round fish days: "But we are both fishes."
Sam thinks and says, "You are right. Let's be friends." They become good friends.

Alla [95]3 years ago

6 0

Answer:

^LOL^

Step-by-step explanation:

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Question 2(Multiple Choice Worth 4 points) (08.02)A pair of equations is shown below. x + y = 5 y = one halfx + 2 If the two equ

8_murik_8 [283]

Answer:

$(2,3)$

Step-by-step explanation:

The first equation is $x+y=5$

The second equation is $y=\frac{1}{2}x+2$

When we graph these two equations, <em>they will meet at a point which represent the solution of the two equations</em>.

We can solve the two equations simultaneously to determine their point of intersection.

Let us substitute the second equation into the first equation to get;

$x+\frac{1}{2}x+2=5$

Multiply through by 2 to get;

$2x+x+4=10$

Group similar terms to obtain;

$2x+x=10-4$

Simplify;

$3x=6$

Divide both sides by 3;

$\Rightarrow x=2$

Put $x=2$ into the second equation;

$y=\frac{1}{2}(2)+2$

$\Rightarrow y=1+2$

$\Rightarrow y=3$

Therefore the graphs of the two functions intersect at (2,3)

See graph in attachment.

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4,500 mass. if you multiply 0.45 and 10,000 you'll get 4,500

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Step-by-step explanation:

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3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

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i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.