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Sever21 [200]
3 years ago
8

WILL GIVE BRAINLIEST! Someone please graph these, I can't graph them.

Mathematics
2 answers:
nadezda [96]3 years ago
8 0

Answer:see attached imageStep-by-step explanation:F(x) = 2x + 5 is a linear graph because the exponent on x is 1.  I tell my students that think of graphs having one less turn/corner that the value of the highest exponent.  so since 2x has a exponent of 1, 1-1 =0 so it has no turns or its a straight line.this has a slope of 2 or 2/1 or up 2 and right 1 from the y intercept which is 5so mark 5 on the y axis and a from there go up 2 and right 1 and make another point.  Join these points and you have your graphand g(x) = (x-5)/2  is its inverse and is found:F(x) = 2x + 5 write it this way y = 2x + 5now swap the x and y    x = 2y - 5solve for yx = 2y + 5- 5       -5 x - 5 = 2y/2        /2 (x-5)/2 = yit can be written as y = x/2 - 5/2and graphed the same way as above with a  1/2 slope and -5/2 y intercept

nadezda [96]3 years ago
4 0

Answer:

see attached image

Step-by-step explanation:

F(x) = 2x + 5 is a linear graph because the exponent on x is 1.  I tell my students that think of graphs having one less turn/corner that the value of the highest exponent.  so since 2x has a exponent of 1, 1-1 =0 so it has no turns or its a straight line.

this has a slope of 2 or 2/1 or up 2 and right 1 from the y intercept which is 5

so mark 5 on the y axis and a from there go up 2 and right 1 and make another point.  Join these points and you have your graph


and


g(x) = (x-5)/2  is its inverse and is found:

F(x) = 2x + 5 write it this way y = 2x + 5

now swap the x and y    x = 2y - 5

solve for y

x = 2y + 5

- 5       -5

x - 5 = 2y

/2        /2

(x-5)/2 = y

it can be written as y = x/2 - 5/2

and graphed the same way as above with a  1/2 slope and -5/2 y intercept

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Let R be the triangular region in the first quadrant with vertices at points (0,0), (a,0), and (0,b), where a and b are positive
Salsk061 [2.6K]

Answer:

volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( \frac{-b}{a}x + b )² dx

Step-by-step explanation:

Given the data in the question and as illustrated in the image below;

R is in the region first quadrant with vertices; 0(0,0), A(a,0) and B(0,b)

from the image;

the equation of AB will be;

y-b / b-0 = x-0 / 0-a

(y-b)(0-a) = (b-0)(x-0)

0 - ay -0 + ba = bx - 0 - 0 + 0

-ay + ba = bx  

ay = -bx + ba

divide through by a

y = \frac{-b}{a}x + ba/a

y = \frac{-b}{a}x + b

so R is bounded by  y = \frac{-b}{a}x + b and y =0, 0 ≤ x ≤ a

The volume of the solid revolving R about x axis is;

dv = Area × thickness

= π( Radius)² dx

= π ( \frac{-b}{a}x + b )² dx

V = π ₀∫^a ( \frac{-b}{a}x + b )² dx

Therefore,  volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( \frac{-b}{a}x + b )² dx

6 0
2 years ago
Can someone please explain how to solve this problem(31)
pogonyaev

Answer:

  (-1.92, 1.08)

Step-by-step explanation:

The <em>incenter</em> is the center of the largest circle that can be inscribed in the triangle. That circle is called the <em>incircle</em>. The incenter is at the point of intersection of the angle bisectors. For a right triangle, the <em>inradius</em> (the radius of the incircle), is found from a simple formula:

  r = (a + b - c)/2 . . . . . where c is the hypotenuse, and a and b are the legs

In your triangle, the inradius is ...

  r = (5 + 3 -√(5² +3²))/2 = 4 -√8.5 ≈ 1.08452

Among other things, this means the coordinates of the incenter are about (1.08, 1.08) from the right angle vertex, so are about ...

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4 0
3 years ago
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marshall27 [118]

Answer:

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Step-by-step explanation:

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