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3 years ago

Find the radius for a circle defined by (x - 1)2 + y2 = 81.

Mathematics

1 answer:

Schach [20]3 years ago

3 0

The radius is 9
The equation is
(x-h)^2 + (y-k)^2 = r^2

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Find P(4).<br> A 1/8<br> B 1/4<br> C 1/2

olga nikolaevna [1]

The correct answer would be A. There is only a 1/8th percent chance of the spinner landing on 4.

4 0

3 years ago

Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4

steposvetlana [31]

$f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}$

$\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}$

6 0

3 years ago

Read 2 more answers

Hi can someone help me with these? 5 and 6 only though

cestrela7 [59]

Answer:

5) 2z - 15 = 9

2z.=9+15

2z =24

z =12

6) 4x-2 = 62°

4x=62°+2°

4x=64°

x=16

6 0

2 years ago

Read 2 more answers

The exterior angle of a certain regular polygon is 60°. How many sides does the polygon have?.

givi [52]

<h3>Answer: 6</h3>

Work Shown:

E = exterior angle = 60 degrees

n = number of sides of the regular polygon

n = 360/E

n = 360/60

n = 6

The regular polygon has 6 sides.

3 0

2 years ago

Determine the solutions of the equation:

jekas [21]

The solutions of the equation are x = 15 and x = -20

<h3>How to determine the solutions of the equation?</h3>

The equation is given as:

the absolute value quantity two fifths times x plus 1 end quantity minus 7 equals 0

Rewrite the equation properly as:

|2/5x + 1| - 7 = 0

Add 7 to both sides

|2/5x + 1| = 7

Remove the absolute bracket

2/5x + 1 = 7 and 2/5x + 1 = -7

Subtract 1 from both sides

2/5x = 6 and 2/5x = -8

Solve for x

x = 15 and x = -20

Hence, the solutions of the equation are x = 15 and x = -20