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GREYUIT [131]
2 years ago
7

Write the equation of the line that passes through the following points. (3, 1) (1,7)

Mathematics
1 answer:
Marina CMI [18]2 years ago
4 0

Answer:

y=-3x+10

Step-by-step explanation:

Slope formula y2-y1/x2-x1

7-1/1-3

simplify

6/-2

simplify

-3

now plug -3 into y=mx+b

y=-3(x)+b

choose any point and plug it into the equation, I am using (3,1)

1=-3(3)+b

simplify

1=-9+b

add 9 to both sides

b=10

plug into slope-intercept form

y=-3x+10

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Answer:

The answer is x = 2.

Step-by-step explanation:

7x + 3 = 2x + 13

  5x + 3 = 13

     5x = 10

       x = 2

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Find the distance between the two points rounding to the nearest tenth (if necessary).
leonid [27]

Answer:

\boxed {\boxed {\sf d\approx 4.2}}

Step-by-step explanation:

The formula for distance is:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Where (x₁, y) and (x₂, y₂) are the points.

We are given (-6, 6) and (-3, 3). If we match the value and its corresponding variable, we see that:

  • x₁= -6
  • y₁ = 6
  • x₂ = -3
  • y₂ = 3

Substitute the values into the formula.

d= \sqrt{ (-3 - -6)^2+(3-6)^2

Solve inside the parentheses.

  • -3 --6 = -3+6 = 3
  • 3-6 = -3

d= \sqrt{(3)^2+ (-3)^2

Solve the exponents.

  • (3)²= 3*3= 9
  • (-3)²= -3*-3 =9

d= \sqrt{9+9

Add.

d= \sqrt18

d= 4.24264069

Round to the nearest tenth. The 4 in the hundredth place tells us to leave the 2 in the tenth place.

d \approx 4.2

The distance between the two points is apprximately <u>4.2</u>

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Im pretty sure
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Set up the integral that uses the method of cylindrical shells to find the volume V of the solid obtained by rotating the region
Ganezh [65]

Answer:

first exercise

V = 16 π

second exercise

V= 68π/15

Step-by-step explanation:

Initially, we have to plot the graph x = (y − 5)2 rotating around y = 3 and the limitation x = 4

<em>vide</em> picture 1

The rotation of x = (y − 5)2 intersecting the plane xy results in two graphs, which are represented by the graphs red and blue. The blue is function x = (y − 5)2. The red is the rotated cross section around y=3 of the previous graph. Naturally, the distance of "y" values of the rotated equation is the diameter of the rotation around y=3 and, by consequence,  this new red equation is defined by x = (y − 1)2.

Now, we have two equations.

x = (y − 5)2

xm = (y − 1)2 (Rotated graph in red on the figure)

The volume limited by the two functions in the 1 → 5 interval on y axis represents a volume which has to be excluded from the volume of the 5 → 7 on y axis interval integration.

Having said that, we have two volumes to calculate, the volume to be excluded (Ve) and the volume of the interval 5 → 7 called as V. The difference of V - Ve is equal to the total volume Vt.

(1) Vt = V - Ve

Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:

(2) V=\int\limits^{y_{1} }_{y_{2}}{2*pi*y*f(y)} \, dy

f(y) represents the radius of the infinitesimal cylinder.

Replacing (2) in (1), we have

V= \int\limits^{5 }_{{3}}{2*pi*y*(y-5)^{2} } \, dy - \int\limits^{7 }_{{5}}{2*pi*y*(y-5)^{2} } \, dy

V = 16 π

----

Second part

Initially, we have to plot the graph y = x2 and x = y2, the area intersected by both is rotated around y = −7. On the second image you can find the representation.

<em>vide</em> picture 2

As the previous exercise, the exclusion zone volume and the volume to be considered will be defined by the interval from x=0 and y=0, to the intersection of this two equations, when x=1 and y = 1.

The interval integration of equation y = x2 will define the exclusion zone. By the other hand, the same interval on the equation x=y2 will be considered.

Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:

(3) V=\int\limits^{x_{1} }_{x_{2}}{2*pi*x*f(x)} \, dx

Notice that in the equation above, x and y are switched to facilitate the calculation. f(x) is the radius of the infinitesimal cylinder

Having this in mind, the infinitesimal radius of equation (3) is defined by f(x) + radius of the revolution, which is 7. The volume seeked is the volume defined by the y = x2 minus the volume defined by x=y2. As follows:

V= \int\limits^{1 }_{{0}}{2*pi*x*(\sqrt{x} + 7) } \, dy - \int\limits^{1 }_{{0}}{2*pi*x*(x^{2}+7) } \, dy

V= 68π/15

6 0
3 years ago
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