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Vedmedyk [2.9K]

3 years ago

5

8 times the sum of a number and 5

1 answer:

kherson [118]3 years ago

4 0

The equation is 8(n+5)

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Evaluate<br><br> a. 6!<br><br> b. 8P5<br><br> c. 12C4

erik [133]

I'm pretty sure that the answer is 12C4 = (12p4)/4! = (12*11*10*9)/4*3*2*1 = 495

Hope this helps

4 0

3 years ago

One side of a square flower bed is 21 feet long. How many plants are needed if they are to be spaced 21 inches apart around the

Lapatulllka [165]

Answer:

48 plants.

Step-by-step explanation:

On one side there would be 21 * 12 inches / 21 inches =- 12 plants.

There are 4 sides so that makes 48 plants.

4 0

3 years ago

PLEASE HELP ME ANSWER THIS BECAUSE IM NOT SMART

dexar [7]

Answer:

negative

Step-by-step explanation:

5 0

3 years ago

lara [203]

21 I hope this helps :)

5 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