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3 years ago

...............................

Mathematics

2 answers:

Otrada [13]3 years ago

6 0

Is something wrong do you need any help on any questions

solong [7]3 years ago

3 0

What is it? Is something wrong

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write and solve an equation to determine the length of a rectangular prism with a width of 4 feet ,a height of 6 feet ,and a vol

Tasya [4]

Answer:

3.5ft

Step-by-step explanation:

To calculate the volume of something, you need to know the formula. Length x width x height. You already know the width, and the height, so you need to find the length.

4ft x 6ft x 3.5ft = 84ft^3

3 0

2 years ago

The person who answes my question can u stay because i have more thankyou

castortr0y [4]

Answer:

c. 60 mins

Step-by-step explanation:

12 * 5 = 60

15 * 4 = 60

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2 years ago

Escribe la expresión para "El producto de 5 y z".

bekas [8.4K]

Answer:

5yz

Step-by-step explanation:

You multiply 5 by y and z

8 0

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

sp2606 [1]

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.

5 0

2 years ago

its time for everybody's favorite subject maaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

zvonat [6]

Answer:

Step-by-step explanation:

Yessir I love math. I know you were being sarcastic but I love math. just multiply everything and you’ll get the answer

4 0

2 years ago

Read 2 more answers