19. Tracey stopped at the convenience store on her way to school. With tax, bottles of soda

A new grocery store is expected to have a fair market value of $2,856,000. It will be assessed at 40%, and the tax rate is $5.24

olganol [36]

Assessed Value = (Fair Market Value) x (0.40) Where 0.40 is the decimal equivalent of 40%. Tax Rate = $5.24/$100 of assessed value = $0.0524 per dollar of assessed value. Taxes = (Assessed Value) x (Tax Rate) = (Assessed Value) x ($0.0524) Hope this helps!

4 0

3 years ago

Read 2 more answers

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

List the numbers which are factors of both 60 and 96

Elden [556K]

The answer is 2. I hope this helped.

6 0

3 years ago

Find the missing term and the missing coefficient 6a −( )5a =( ) a2 − 35a

andrew11 [14]

Solving for the missing term and the missing coefficient (6a − )5a = ( ) a^2 − 35a
Let the missing term be X
Let the missing coefficient be Y
Therefore, (6a – X)5a = Y(a^2) – 35a
6a x 5a – X.5a = Y.a^2 – 35a
30a^2 – X.5a = Y.a^2 – 35a
Equating co-efficients,
30a^2 = Y.a^2; X.5a = 35a
30 = Y; 5X = 35
Y = 30; X = 7
Therefore, (6a-7)5a = 30 a^2 – 35a <span>
</span>

3 0

3 years ago

If the rule of the function is "multiply by 2 then add 3" and the input value is -2, what is the output value?

miv72 [106K]

The equation would be 2x+3 so when you plug it in it's 2(-2)+3 which simplify to -4+3 which is -1

5 0

3 years ago