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OLEGan [10]
3 years ago
13

Meredith lives 24 blocks from her friends house if she travels on block every minute how many minutes will it take her to reach

her friends house show how you calculate each answer
Mathematics
1 answer:
Anestetic [448]3 years ago
3 0

Meredinth takes 24 minutes to reach her friend house

<em><u>Solution:</u></em>

Given that, Meredith lives 24 blocks from her friends house

She travels one block every minute

<em><u>To find: time taken by Meredith to reach friend house</u></em>

From given,

Number of blocks between Meredith house and her friedn house = 24

She travels one block every minute

Thus she takes 1 minute for 1 block

One block = 1 minute

So, for 24 blocks, we have to multiply by 24

24\ block = 1 \times 24\ minute\\\\24\ block = 24\ minute

Thus Meredinth takes 24 minutes to reach her friend house

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Given the volume is 105 inches³ for an 8 pound bowling ball, find the radius of the bowling ball,
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Step-by-step explanation:

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1 year ago
Is this answer right?
zalisa [80]

Answer:

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Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
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d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
3 years ago
Question 2 2.1 2.2 2.2.1 2.2.2 Determine the general solution for sin(x - 30°) = cos 2x Consider the functions f(x) = sin(x-30°)
OLEGan [10]

The general solution for the given equation is 40°+120°n.

The period of g(x) is π.

The range of f(x) is [-1, 1].

  • sin(x-30°) = cos 2x
  • sin(x-30°) = sin(90°-2x)
  • (x-30°) = (90°-2x) + 360°n
  • x + 2x =  90° + 30° + 360°n
  • 3x = 120° + 360°n
  • x = (120° + 360°n)/3
  • The general solution (x) is 40° + 120°n.
  • g(x) = cos 2x
  • The period of cos x is 2π.
  • If the multiplying factor of x is 'n', then it decreases the period by n times.
  • Here, in g(x), n is 2.
  • The period of g(x) is equal to half the period of cos x.
  • The period of g(x) = 2π/2
  • The period of g(x) is π.
  • f(x) = sin(x-30°)
  • "Sin" is a trigonometric function which has a range from -1 to 1.

To learn more about  period, visit :

brainly.com/question/12539110

#SPJ9

5 0
2 years ago
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