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Graph of a Circle: (x - h)^2 + (y - k)^2 = r^2

Circle with center (h, k), radius r. 

Move variable terms to one side:

\(x^2-16x+y^2+22y=86\)

Complete the square:

\((x-8)^2-64+(y+11)^2-121=86\)

\((x - 8)^2+(y+11)^2=86+121+64=271=r^2\)

The area of the circle is \(\pi{r^2}=271\pi\)

Combine like terms:

y = -5x^2 + 15x - 10

Vertex form: a(x - h)^2 + k

Standard form: ax^2 + bx + c

a = -5, b = 15, c = -10

Vertex form: -5(x - b/2a)^2 + k = -5(x - 3/2)^2 + 45/4 - 10 = -5(x - 3/2)^2 + 5/4.

Vertex = (h, k) = (3/2, 5/4)

Let the square base have side length a. The area of the triangle with slant height b has area ab/2 = (a^2)/3, so 2a/3 = b. 

Thus, the area of each surface triangle is (a^2)/3, we have four of them with a base of a^2, so the surface area = (7a^2)/3 = 432

Solving for a, you get a = 36/sqrt(7)

The height of the pyramid, using pythagorean theorem with legs h and a/2 with hypotenuse 2a/3 is:

\(({2a\over{3}})^2-({a\over{2}})^2=h^2\)

Substitution:

\(({24\over{\sqrt{7}}})^2-({18\over{\sqrt{7}}})^2={576\over{7}}-{324\over{7}}=36=h^2\)

Solving for h, we get h = 6.

The volume of a pyramid, being base area * height / 3 would be a^2 * 6 / 3 = 1296*2/7 = 2592/7

x - y^2 + 8y = 13 + 2y + y^2 - 7y - 14, combining like terms:

x = 2y^2 - 13y - 1

Note that this is a function of y, so f(y) = x, which will look like a sideways parabola when graphed. Still, the vertex can be calculated in the same manner as if it was f(x) = y.

Vertex form can be written as a(x - h)^2 + k = 0, where (h, k) is the vertex. In this problem specifically, the x = y, and the variables can be solved for using "complete the square". However, note that for our problem specifically, the vertex would be (k, h) because the this parabola is the inverse of a(x - h)^2 + k = 0.

2(y - 13/4)^2 - 169/8 - 1 = 2(y - 13/4)^2 - 177/8 = 0 

(h, k) = (13/4, -177/8), so the vertex to our problem, (k, h) = (-177/8, 13/4)

In triangle ABC, angle BAC and angle ABC both measure 60 degrees. Because the interior angle sum of a triangle adds up to 180 degrees, our third angle BCA must equal 60 degrees... and then since all three angles are the same, we have an equilateral triangle ABC.

Because ABC is equilateral, then the angle bisector of BAC also acts as the perpendicular height of the triangle = 24.

Angle ABC = 60 degrees, angle BAD = 60/2 = 30 degrees, and angle ADB = 90 degrees, so we have a 30 - 60 - 90 triangle with long leg of 24 units, thus the shorter leg BD would have length 24/sqrt(3) = 8sqrt(3) after rationalizing the denominator. BC = 2*BD because the height of the isosceles triangle also bisects the base.

Base * Height / 2 = Area[ABC] = 24*16sqrt(3)/2 = \(192\sqrt{3}\) units\(^2\).

ax + by = c

a(1) + b(7) = c

a + 7b = c

a(2) + b(-3) = c

2a - 3b = c

Transitive property:

a + 7b = 2a - 3b

a = 10b.

WLOG, set a = 10, then b = 1.

10x + 1y = c

10(1) + 1(7) = c

c = 17

Thus, your equation ax + by = c is 10x + y = 17

x^2 + 7x - 4 = -x^2 + 9x + 6, combine like terms:

2x^2 - 2x - 10 = 0

x^2 - x - 5 = 0

By vieta's, the sum of the roots a + b = -1 * the second term / the leading coefficient = 1.

Thus, a + 3 + b + 3 = a + b + 6 = 7

I = (4,r)   where r is the radius of the incircle

To find r

[ ABC] = (1/2)(8 + 5 + 5) r

(1/2) 8 * 3 = (1/2) (18) r

r = 4/3

l = (4, 4/3)

ID = 4/3 

AI =  3 - (4/3)  = 5/3

Triangles  AMN  and ABC are similar such that  AI /  AD =   = (5/3) / 3  = 5/9

[AMN ] = (5/9)^2   [ ABC  ] =   (5/9)^2 (1/2) (8)(3)  = (25 / 81 ) * 12  = 300 / 81 =   100 / 27

Area = 

(1/2) ( 6)(8) sin 35°   ≈  13.77

Here :  https://web2.0calc.com/questions/painting-a-wall

Rate * Time   = Distance

Let T be the normal time in hours

25 minutes =  25/60  = 5/12 hr

Equate distances

40 T  =  55 ( T - 5/12)

40T  = 55T  - 275 /  12

275  / 12  =  55T - 40T

275 / 12  = 15  T

275 / [ 12 * 15 ]   = T

55/36 hrs  = T

D =  40 (55 / 36)  ≈  61.1  miles

A

4              5

B   3-x      D     x        C

             3

Since AD bisects  BAC

Let  BD  = 3 - x

Let  CD =  x

So, since AD is a  bisector,   we have that

BD / AB  = CD / AC

(3 - x) / 4  = x / 5

5 (3 - x)  = 4x

15 - 5x = 4x

15 = 9x

x = 15/9  = 5/3 =  CD

[ ADC ] = (1/2) (CD) ( AB)  =  (1/2) ( 5/3) ( 4)  =   10 / 3  =  3 { to the nearwst integer }

Here is a  similar problem

 https://web2.0calc.com/questions/geometry_55509

Here : https://web2.0calc.com/questions/rates_51

Let the side of the square  =   x

CD =  1  - x

FB = 3 - x

Triangles  CDE  and  EFB  are similar

CD / DE  = EF / FB

(1 - x) / x  = x / (3 - x)

x^2  = (1 - x) ( 3 - x)

x^2  = x^2 - 4x + 3

4x = 3

x  = 3/4

Area of square  =  (3/4)^2  = 9 /16

Notice  that the area =  [ product of legs / sum of legs ] ^2  =  [ (1 *3) / ( 1 + 3) ] ^2  = (3 / 4)^2 =  9 / 16

CD = BD   = 10

ED = ED

Thus by LL , triangle CDE  is  congruent to triangle BDE

So CE = BE

sin CED / CD  = sin CDE / CE

sin 75 / 10  = sin 90 / CE

CE  =  10 sin 75  =  BE ≈  10.35

x^2  + y^2  + 2x - 4y + 8 + 10x  -12y

x^2 + 12x + y^2 - 16y  +  8         

Take the derivative with respect to x and  set to 0

2x + 12  =  0    →   x + 6 = 0  →  x  = -6

Take the derivative with respect  to y and  set to  0

2y  - 16  =  0   →  y  -  8  = 0  →   y =  8

The minimum occurs at   (-6,8)

And the minimum is

(-6)^2 + 12 (-6) + (8)^2  - 16(8) + 8 =    -92

Since CX  is a bisector  and AC = BC, then AX = BX

Then ACX and BCX   are congruent  so  their areas  are   1/1

radius =  8*15*12  /  4*Area

radius = 1440  /  [ 4 *   sqrt  [ 17.5 (17.5 - 8)(17.5 - 15)(17.5 -12) ]  ]  ≈  7.53 

a^3 - b^3    = 1

a      b         ab      a + b

1      0          0           1

0     -1          0          -1

Slope of line =  (-7-2) / ( 4 - -8)  = -9 / 12  =  -3/4

Using the slope and ( 4, -7)   the equation of the line is

y = (-3/4) ( x - 4)  - 7

y = -3/4x + 3  - 7         mulriply through by 4

4y = -3x - 16

3x + 4y = -16

Simplify L1

-3y = -11x - 14

y = (11/3)x - 14

The slope of a perpendicular line is -3/11

Using (5, -6), the equation of the perpendicular line L2  is

y = (-3/11)(x - 5)   -6

y = (-3/11)x + 15/11 - 66/11

y = (-3/11)x - 51/11

We can use similar triangles here. First of all, ABC is an isosceles triangle because the altitude bisects the base, and thus forms two congruent triangles on each side by SAS congruency...

Angle BAD = angle FCB because both angles = 180 - 90 - angle B through angle-chasing. Ang le BAD = angle DAC, so we now can form a general similarity statement:

Triangle HCD ~ Triangle CAD. This means that AD / CD = CD / HD.

Let CD = s: \({16\over{s}} = {s\over{5}} \), and then cross multiplying yields \(80 = s^2\), square rooting both sides gives \(s = 4\sqrt{5}\).

Because the area of ABC = bh/2 = AD * BC/2 = AD*DC, we can substitute = 16 * 4sqrt(5) = 64sqrt(5), our final answer.

Looking at this problem, we don't see we have that many options, so we use our backup plan for any geometry problem: trigonometry. 

Set AE as x. The side length of triangle DEF is 10, so BF = 7, and AF = 10-x.

We want to write some equation, with x, so we look at the triangle in the middle, BCA, and hope to use the pythagorean theorem.

Now in each of these triangles, BDC, AEC, BFA, we have 2 sides and an angle, and we want to find the side opposite to the angle.

Luckily, for us we have the Law of Cosines!

\({BC}^{2}=9+49-2*3*7*\cos({60}^{\circ})\). Simplifying, we get, \({BC}^{2}=37\). 

Similarly, \({AC}^2=9+{x}^{2}-2*3*x*\cos({60}^{\circ})\). Simplifying, we, get \({AC}^{2}=9+{x}^{2}-3x\).

Also, \({AB}^2=49+{(x-10)}^{2}-2*7*(x-10)*cos({60}^{\circ})\). Simplifying, we get, \({AB}^{2}={x}^{2}-27x+219\).

Using pythagorean theorem, we have \(37+{x}^2-3x+9={x}^{2}-27x+219\). Solving and cancelling, we get \(x=\frac{173}{24}\). Our answer is 173/24.

\(\)

Let the distance on the base line, from the right edge to the strip, be x.

Then the following applies:

\( A_{rectangle}=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0 \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ A_{strip}=1\cdot \sqrt{8.5375^2+8^2}\\ \color{blue}A_{strip}=11.700\)

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