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3 years ago

6

Can someone answer these questions and explain the reasoning?

1 answer:

marishachu [46]3 years ago

7 0

Answer:

idk number 6

Step-by-step explanation:

I hope this is what you need

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Maria lives 98 miles away from her grandparents. She wants to be at her grandparents’ house by 3:00 pm. If she leaves her house

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b

Step-by-step explanation:

b

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3 years ago

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This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

pashok25 [27]

Answer with Step-by-step explanation:

We are given that

$f(x,y,z)=10x+10y+3z$

$g(x,y,z)=5x^2+5y^2+3z^2=43$

We have to find the extreme values of the function.

$f_x(x,y,z)=10,f_y(x,y,z)=10,f_z(x,y,z)=3$

$g_x(x,y,z)=10x,g_y(x,y,z)=10y,g_z(x,y,z)=6z$

$\Delta f(x,y,z)=\lambda \Delta g(x,y,z)$

Therefore,

$10=\lambda 10x\implies x=\frac{1}{\lambda}$

$10=10y\lambda\implies y=\frac{1}{\lambda}$

$3=6z\lambda$

$z=\frac{1}{2\lambda}$

Substitute the values in g(x)

Then, we get

$\frac{5}{\lambda^2}+\frac{5}{\lambda^2}+\frac{3}{4\lambda^2}=43$

$\frac{20+20+3}{4\lambda^2}=43$

$\frac{43}{4\lambda^2}=43$

$\lambda^2=\frac{43}{4\times 43}$

$\lambda=\pm\frac{1}{2}$

When $\lambda=\frac{1}{2}$ then,

$x=2,y=2,z=1$

When $\lambda=-\frac{1}{2}$

Then, $x=-2,y=-2,z=-1$

$f(2,2,1)=10+10+3=23$

$f(-2,-2,-1)=-10-10-3=-23$

Hence, maximum value of f(x,y,z) is 23 at (2,2,1) and minimum value of f(x,y,z) is -23 at (-2,-2 ,-1).

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4 years ago

9. Zane owes his dad money for his cell phone that he

ozzi

Answer:

T=45m+23

Step-by-step explanation:

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3 years ago

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The number of bacteria p in a pond after t hours is p(t). the more bacteria there are, them more rapidly the number increase. Th

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In a pond after t hours is p(t). The more bacteria there are, the more rapidly the number increases. This is expressed by the equation p'(t)= 2p(t) . Find.

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3 years ago

WILL GIVE BRAINLIEST!!!!

Vesna [10]

$(2^8 \times 5^{-5} \times 19^0)^{-2} \times (\dfrac{5^{-2}}{2^3})^4 \times 2^{28} =$

$= 2^{-16} \times 5^{10} \times 1 \times \dfrac{5^{-8}}{2^{12}} \times 2^{28}$

$= 2^{-16} \times 5^{10} \times 5^{-8} \times 2^{-12} \times 2^{28}$

$= 2^{-16} \times 2^{-12} \times 2^{28} \times 5^{10} \times 5^{-8}$

$= 2^{-16-12+28} \times 5^{10-8}$

$= 2^{0} \times 5^{2}$

$= 1 \times 25$

$= 25$

7 0

3 years ago