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____ [38]
3 years ago
13

"a man drove 11 mi directly east from his home, made a left turn at an intersection, and then traveled 2 mi north to his place o

f work. if a road was made directly from his home to his place of work, what would its distance be to the nearest tenth of a mile?"
Mathematics
1 answer:
Charra [1.4K]3 years ago
8 0
Pythagorean therom!


east and the north make a right angle

a^2 + b^2 = c^2
(11)^2 + (2)^2 = c^2
121 + 4 = c^2
125 = c^2
11.2 miles = c
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A thin metal plate, located in the xy-plane, has temperature T(x, y) at the point (x, y). Sketch some level curves (isothermals)
Sophie [7]

Answer:

Step-by-step explanation:

Given that:

T(x,y) = \dfrac{100}{1+x^2+y^2}

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)

By cross multiply

c({1+x^2+2y^2}) = 100

1+x^2+2y^2 = \dfrac{100}{c}

x^2+2y^2 = \dfrac{100}{c} - 1 \ \  -- (2)

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1

This is the equation for the  family of the eclipses centred at (0,0) is :

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

a^2 = \dfrac{100}{c} -1  \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}

Therefore; the level of the curves are all the eclipses with the major axis:

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The sketch of the level curves can be see in the attached image below.

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3 years ago
Solve the equation 3x+5y=15 for y
larisa86 [58]
<span> 3x+5y = 15
       5y = -3x + 15
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hope it helps


</span>
7 0
3 years ago
What is 3 tenths greater than 80%
elena-14-01-66 [18.8K]

we know that


3 tenths is equal to

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so


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0.30+0.80=1.10

therefore


the answer is

1.10

6 0
3 years ago
Read 2 more answers
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