Ask question

Ask question

All categories

sleet_krkn [62]

3 years ago

8

Help meh plz plz quickly

1 answer:

Dahasolnce [82]3 years ago

6 0

1 load of stone equals 1/6 ton

You might be interested in

5 < 2a(üssü -1) + 3 < 9 eşitsizliğini sağlayan a sayısı aşağıdakilerden hangisi olabilir?

Andru [333]

Answer:

d

Step-by-step explanation:

I believe it is d if I'm not wrong

6 0

2 years ago

Rewrite the expression as a single logarithm: <br> log 5 - log 7

Lady bird [3.3K]

Answer:

-0.146

Step-by-step explanation:

log 5 = 0.699

log 7 = 0.845

5 0

3 years ago

Read 2 more answers

Area of octagon with radius 5?

Montano1993 [528]

I got 200 from a octagon with the diameter of 10

3 0

2 years ago

WILL MARK THE BRAINLIEST

Elis [28]

Answer:

Option D, 16/21

Step-by-step explanation:

<u>Step 1: Add all of them together</u>

5 + 7 + 9

21

<u>Step 2: Figure out how much of the plants are medicinal</u>

7 + 9

16

<u>Step 3: Figure out the probability </u>

16/21

Answer: Option D, 16/21

7 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

3 years ago