Answer:
[tex]\textsf{13.} \quad(x-1)^2+(y+2)^2=9[/tex]
center = (1, -2)
radius = 3
[tex]\textsf{14.} \quad y=5 \cdot 3^x[/tex]
[tex]\textsf{15.} \quad x=\dfrac{104}{3}, \quad x=-\dfrac{88}{3}[/tex]
Step-by-step explanation:
Question 13
Equation of a circle
[tex](x-a)^2+(y-b)^2=r^2[/tex]
(where (a, b) is the center and r is the radius)
Given equation:
[tex]x^2+y^2-2x+4y-4=0[/tex]
To rewrite the given equation in graphing form, first add 4 to both sides of the equation and rearrange the variables:
[tex]\implies x^2-2x+y^2+4y=4[/tex]
Add the square of half the coefficients of the x and y terms to both sides:
[tex]\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[/tex]
[tex]\implies x^2-2x+1+y^2+4y+4=9[/tex]
Finally, factor both trinomials:
[tex]\implies (x-1)^2+(y+2)^2=9[/tex]
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:
[tex]y=ab^x[/tex]
(where a and b are constants to be found)
Given points on the curve:
Substitute the given points into the general form of the equation to create two equations:
[tex]\textsf{Equation 1}: \quad ab^4=405[/tex]
[tex]\textsf{Equation 2}: \quad ab^9=98415[/tex]
Divide Equation 2 by Equation 1 to eliminate a, then solve for b:
[tex]\implies \dfrac{ab^9}{ab^4}=\dfrac{98415}{405}[/tex]
[tex]\implies \dfrac{b^9}{b^4}=243[/tex]
[tex]\implies b^{9-4}=243[/tex]
[tex]\implies b^5=243[/tex]
[tex]\implies b=\sqrt[5]{243}[/tex]
[tex]\implies b=3[/tex]
Substitute the found value of b into one of the equations and solve for a:
[tex]\implies a(3)^4=405[/tex]
[tex]\implies 81a=405[/tex]
[tex]\implies a=\dfrac{405}{81}[/tex]
[tex]\implies a=5[/tex]
Therefore, the exponential equation that passes through (4, 405) and (9, 98415) is:
[tex]y=5 \cdot 3^x[/tex]
Question 15
Given equation:
[tex]\left|\dfrac{3}{2}x-4\right|+2=50[/tex]
Isolate the absolute value by subtracting 2 from both sides:
[tex]\implies \left|\dfrac{3}{2}x-4\right|=48[/tex]
Set the contents of the absolute value to positive and negative and solve for x:
Positive absolute value:
[tex]\implies \dfrac{3}{2}x-4=48[/tex]
[tex]\implies \dfrac{3}{2}x=52[/tex]
[tex]\implies x=52 \cdot \dfrac{2}{3}[/tex]
[tex]\implies x=\dfrac{104}{3}[/tex]
Negative absolute value:
[tex]\implies -\left(\dfrac{3}{2}x-4\right)=48[/tex]
[tex]\implies -\dfrac{3}{2}x+4=48[/tex]
[tex]\implies -\dfrac{3}{2}x=44[/tex]
[tex]\implies x=44 \cdot -\dfrac{2}{3}[/tex]
[tex]\implies x=-\dfrac{88}{3}[/tex]
Therefore, the solution to the equation is:
[tex]x=\dfrac{104}{3}, \quad x=-\dfrac{88}{3}[/tex]
