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son4ous [18]
3 years ago
9

Henrietta has 120 to buy a new bicycle.She finds the bike that she wants to buy on sale for 20% off. If there is 6% sales tax, w

hich of the following best describes the amount of money Henrietta has with regards to buying the bike?a. Henrietta has enough before the tax, but not enough after the tax. b. Henrietta has enough money even after the tax. c. Henrietta does not have enough money even before the tax. d. There is not enough information given.
Mathematics
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

D

Step-by-step explanation:

In this question, option D is the correct answer. Firstly, we were made to know the amount of money she has to be $120. Now, to properly account for the 20% discount and also the 6% after sales tax, we need to know actually the amount the bike costs.

If we are furnished with the information about what the bike cost, we can then proceed to adequately find the values of the discount and the value of what the after sales tax would be. It is after this that we can now make any comparison with the money she has at hand and thus consider other options in the question.

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A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 1
Jet001 [13]

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth= \$ 17,000

depreciates by \frac{1}{2} in 3 years.

time=  19 years

depreciates rate=?

Using formula:

\to \text{Worth=  Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}

\to A_t=A_0(1-\frac{r}{100})^t

calculates depreciate value in 3 year = \frac{1}{2} \times 17,000

                                                              = 8,500

so,

A_t=8,500\\\\A_0=17,000\\\\t=3\ years

\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%

depreciates rate= 20%

\to \text{Worth=  Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}

= \$ 17,000 (1- \frac{20}{100})^{19}\\\\= \$ 17,000 (1-0.2)^{19}\\\\= \$ 17,000 (0.8)^{19}\\\\= \$ 17,000 \times 0.014\\\\= \$ 238

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The Statistical significance provides a link between the  dependent and the independent variable. It provides us the indication that the change in the independent variable is correleated with the shift caused in the dependent variable.

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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}

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1 year ago
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