Question

NO LINKS!!!! Please assist me with these last 3 problems​

Answer 1

`\textsf{13.} \quad(x-1)^2+(y+2)^2=9`

center = (1, -2)

radius = 3

`\textsf{14.} \quad y=5 \cdot 3^x`

`\textsf{15.} \quad x=(104)/(3), \quad x=-(88)/(3)`

Explanation:

Question 13

Equation of a circle

`(x-a)^2+(y-b)^2=r^2`

(where (a, b) is the center and r is the radius)

Given equation:

`x^2+y^2-2x+4y-4=0`

To rewrite the given equation in graphing form, first add 4 to both sides of the equation and rearrange the variables:

`\implies x^2-2x+y^2+4y=4`

Add the square of half the coefficients of the x and y terms to both sides:

`\implies x^2-2x+\left((-2)/(2)\right)^2+y^2+4y+\left((4)/(2)\right)^2=4+\left((-2)/(2)\right)^2+\left((4)/(2)\right)^2`

`\implies x^2-2x+1+y^2+4y+4=9`

Finally, factor both trinomials:

`\implies (x-1)^2+(y+2)^2=9`

Now compare the equation in graphing form to the general equation of a circle to find the center and radius:

⇒ a = 1 and b = -2 ⇒ center = (1, -2)

⇒ r² = 9 ⇒ radius = √9 = 3

Question 14

General form of an exponential function:

`y=ab^x`

(where a and b are constants to be found)

Given points on the curve:

• (4, 405)
• (9, 98415)

Substitute the given points into the general form of the equation to create two equations:

`\textsf{Equation 1}: \quad ab^4=405`

`\textsf{Equation 2}: \quad ab^9=98415`

Divide Equation 2 by Equation 1 to eliminate a, then solve for b:

`\implies (ab^9)/(ab^4)=(98415)/(405)`

`\implies (b^9)/(b^4)=243`

`\implies b^(9-4)=243`

`\implies b^5=243`

`\implies b=\sqrt[5]{243}`

`\implies b=3`

Substitute the found value of b into one of the equations and solve for a:

`\implies a(3)^4=405`

`\implies 81a=405`

`\implies a=(405)/(81)`

`\implies a=5`

Therefore, the exponential equation that passes through (4, 405) and (9, 98415) is:

`y=5 \cdot 3^x`

Question 15

Given equation:

`\left|(3)/(2)x-4\right|+2=50`

Isolate the absolute value by subtracting 2 from both sides:

`\implies \left|(3)/(2)x-4\right|=48`

Set the contents of the absolute value to positive and negative and solve for x:

Positive absolute value:

`\implies (3)/(2)x-4=48`

`\implies (3)/(2)x=52`

`\implies x=52 \cdot (2)/(3)`

`\implies x=(104)/(3)`

Negative absolute value:

`\implies -\left((3)/(2)x-4\right)=48`

`\implies -(3)/(2)x+4=48`

`\implies -(3)/(2)x=44`

`\implies x=44 \cdot -(2)/(3)`

`\implies x=-(88)/(3)`

Therefore, the solution to the equation is:

`x=(104)/(3), \quad x=-(88)/(3)`

Answer 2

Answer + Step-by-step explanation:

13.

x² + y² - 2x + 4y - 4 = 0

(x² - 2x) + (y² + 4y )- 4 = 0

x² - 2x = (x² - 2x + 1) - 1

= (x - 1)² - 1

y² + 4y = (y² + 4y + 4) - 4

= (y + 2)² - 4

Then

x² + y² - 2x + 4y - 4 = 0

⇔ (x - 1)² - 1 + (y + 2)² - 4 - 4 = 0

⇔ (x - 1)² + (y + 2)² = 9

⇔ (x - 1)² + (y + 2)² = 3²

Then

The center is (1 , -2)

The radius is 3

……………………………………………

14.

(4 , 405) ; (9 , 98415)

y = abˣ

The graph passes through (4 , 405) ⇒ 405 = ab⁴

The graph passes through (9 , 98415) ⇒ 98415 = ab⁹

We divide:

ab⁹÷ab⁴ = 98415 ÷ 405

⇔ b⁵ = 243

⇔ b⁵ = 3⁵

⇔ b = 3

Finding a :

405 = ab⁴

⇔ 405 = a3⁴ = 81a

⇔ a = 405/81 = 5

Conclusion:

The exponential equation is :

y = 5×3ˣ

………………………………………………

15.

|3/2x - 4| + 2 = 50

Then

|3/2x - 4| = 48

Then

3/2x - 4 = 48 or 3/2x - 4 = -48

Then

3/2x = 52 or 3/2x = -44

Then

x = 104/3 or x = -88/3