Question

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

Answer 1

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

Answer 2

Answer:

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

       center = (1, -2)

       radius = 3

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

$\textsf{15.} \quad x=\dfrac{104}{3}, \quad x=-\dfrac{88}{3}$

Step-by-step 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:

$\implies x^2-2x+\left(\dfrac{-2}{2}\right)^2+y^2+4y+\left(\dfrac{4}{2}\right)^2=4+\left(\dfrac{-2}{2}\right)^2+\left(\dfrac{4}{2}\right)^2$

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

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

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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:

$\implies \dfrac{ab^9}{ab^4}=\dfrac{98415}{405}$

$\implies \dfrac{b^9}{b^4}=243$

$\implies b^{9-4}=243$

$\implies b^5=243$

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

$\implies b=3$

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

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

$\implies 81a=405$

$\implies a=\dfrac{405}{81}$

$\implies a=5$

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

$y=5 \cdot 3^x$

Question 15

Given equation:

$\left|\dfrac{3}{2}x-4\right|+2=50$

Isolate the absolute value by subtracting 2 from both sides:

$\implies \left|\dfrac{3}{2}x-4\right|=48$

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

Positive absolute value:

$\implies \dfrac{3}{2}x-4=48$

$\implies \dfrac{3}{2}x=52$

$\implies x=52 \cdot \dfrac{2}{3}$

$\implies x=\dfrac{104}{3}$

Negative absolute value:

$\implies -\left(\dfrac{3}{2}x-4\right)=48$

$\implies -\dfrac{3}{2}x+4=48$

$\implies -\dfrac{3}{2}x=44$

$\implies x=44 \cdot -\dfrac{2}{3}$

$\implies x=-\dfrac{88}{3}$

Therefore, the solution to the equation is:

$x=\dfrac{104}{3}, \quad x=-\dfrac{88}{3}$