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Sloan [31]
2 years ago
10

An art gallery displays a large painting in the center of a wall that is 24 feet ( ft wide. The painting is 10ft wide. Which of

the following equations can be used to find the distances, x, in feet, from the left end of the wall to the edges of the painting?

Mathematics
1 answer:
san4es73 [151]2 years ago
8 0

the complete question in the attached figure

step 1

Find the distance of the center of the wall from the left end of the wall

distance=(1/2)*24--------> 12 ft

so

If x is the distance from a painting edge to its corresponding end of the wall, then

|x-12| is the distance from the painting edge to the center of the wall

and

If the painting is centered, then 2|x-12| represents the width of the painting

Since the painting is 10 ft wide,

we have 2|x-12|=10

therefore

the answer is the option B

2|x-12|=10

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Evaluate the integral by making an appropriate change of variables.
VARVARA [1.3K]

By inspecting the integrand, the "obvious" choice for substitution would be

<em>u</em> = <em>y</em> + <em>x</em>

<em>v</em> = <em>y</em> - <em>x</em>

<em />

Solving for <em>x</em> and <em>y</em>, we would have

<em>x</em> = (<em>u</em> - <em>v</em>)/2

<em>y</em> = (<em>u</em> + <em>v</em>)/2

in which case the Jacobian and its determinant are

J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12

The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.

Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So

-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>

-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2

-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>

-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>

<em />

So, the integral becomes

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=\displaystyle\frac52\int_8^9\frac u7(\sin7-\sin(-7))\,\mathrm du

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4 0
2 years ago
Calculate the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass of 1 m^3 of water to be 1000 kg
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Answer:

1813.137 KJ

Step-by-step explanation:

potential energy of the body = mgH

where m is mass in Kg , g= 9.81 m/sec^2 and H= height in m

here m= 1000 kg, g= 9.81 m/s^2  and H= 607 feet = 607×0.305= 185.135 m

hence the potential energy p= 1000×9.81×185.135= 1813137.2 J

= 1813.137 KJ

hence the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass as 1000 kg is = 1813.137 KJ

4 0
3 years ago
A certain right triangle has area 30 in. squared . one leg of the triangle measures 1 in. less than the hypotenuse. let x repres
Brut [27]
A) The length of the longer leg is x-1

b) Based on the area, the other leg is 2*30/(x -1). Based on the Pythagorean theorem, the other leg is √(x^2 -(x -1)^2).

c) Equating the two expressions for the shorter leg, we have
.. 60/(x -1) = √(2x -1)
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.. 2x^3 -5x^2 +4x -3601 = 0

d) There is one positive real root, at x=13. A graphical solution works well.

The three sides of the triangle are 5 in, 12 in, 13 in.


_____
5-12-13 is a well-known Pythagorean triple. It is the next smallest one after 3-4-5.

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Leviafan [203]
Because of the transversal and the parallel lines, we know that A+B=180°.

So:
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A = 90 + 20 = 110

(Let’s check:
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8 0
3 years ago
How to estimate 348?
schepotkina [342]
348 rounded to tens place = 350
348 rounded to hundreds place = 300 (possibly 400)
4 0
3 years ago
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