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aleksandrvk [35]
2 years ago
12

What is the equation of a line that contains point (4, –3) and is PARALLEL to –4x + y = 7?

Mathematics
2 answers:
andrew-mc [135]2 years ago
8 0
Nfjjrjejeejjetyggbhccccccftdurfgo
Otrada [13]2 years ago
4 0

Answer:

(4,3),y=(7)/(4)x+1

Step-by-step explanation:

To find an equation that is parallel to y = 7 4 x + 1 y=74x+1, the slopes must be equal. Using the slope of the equation, find the parallel line using the point-slope formula. ( 4 , 3 ) (4,3) m = 7 4

Using the point-slope form y − y 1 = m ( x − x 1 ) y-y1=m(x-x1), plug in m = 7 4 m=74, x 1 = 4 x1=4, and y 1 = 3 y1=3. y − ( 3 ) = ( 7 4 ) ( x − ( 4 ) )

Solving for y:

Multiply − 1 -1 by 3 3. y − 3 = ( 7 4 ) ( x − ( 4 ) ) y-3=(74)(x-(4)) Simplify ( 7 4 ) ( x − ( 4 ) ) (74)(x-(4)).

y − 3 = 7 x 4 − 7

y = 7 x 4 − 4 y=7x4-4 Reorder terms. y = 7 4 x − 4

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Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
The description for a certain brand of house paint claims a coverage of 460 ft2/gal. (a) express this quantity in square meters
Angelina_Jolie [31]

Given 460 ft^2/gal

a) convert it into m^2/l

=> It means ft^2 -> m^2 and gal -> l

1 ft = 0.3048 m

1 ft^2 = 0.092903 m^2

1 gal = 3.78541 l

460ft^2= [tex]460\frac{ft^2}{gal} = \frac{42.7354m^2}{3.78541l} = 11.2895m^2/l m^2[/tex].

b.  11.2895\frac{m^{2} }{l} *1000\frac{l}{m^3} =11289.5 m^{-1}

c.Given=  460\frac{ft^2}{gal}

To find inverse, divide it by 1.

Inverse- \frac{1}{460} \frac{gal}{ft^2}




6 0
3 years ago
The sum of the angle measures of a triangle is 180°. Find the value of x. Then find the angle measures of the triangle.
MAXImum [283]

Answer:

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  • 43°, 51°, 86°

Step-by-step explanation:

The sum of the angles is ...

  x° + (x +8)° + 2x° = 180°

  4x +8 = 180 . . . . . . . . . . . collect terms, divide by °

  x +2 = 45 . . . . . . . . . . divide by 4

  x = 43 . . . . . . . . . . subtract 2

  x +8 = 51

  2x = 86

The angles are 43°, 51°, 86°.

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2 years ago
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2 years ago
Which statement describes the function shown on the graph
Anna007 [38]
Let’s use process of elimination
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2 years ago
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