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3 years ago

The admission fee at a carnival is 3.00 for children and $5.00 for adults. On the first day 1500 people enter the fair

Mathematics

1 answer:

Natasha2012 [34]3 years ago

8 0

Question:

The admission fee at a carnival is $3.00 for children and $5.00 for adults. On the first day 1,500 people enter the fair and $5740 is collected. How many children and how many adults attended the carnival?

Answer:

Number of children and adults attended the carnival are 880 and 620 respectively

Step-by-step explanation:

Given:

The admission fee at a carnival for children = $3.00

The admission fee at a carnival for adults = $5.00

Number of people entered the fare on first day= 1500

Total amount collected on the first day = $5740

To Find:

Number of children and adults attended the carnival =?

Solution:

Let

The number of children be x

The number of adults be y

The we know that on the first day the total number children visited the fare was 1500

x + y = 1500

x = 1500 - y------------------------------(1)

Also the total fare collected was $5740

x(3) + y(5) = 5740

3x + 5y = 5740-------------------(2)

Substituting (1) in (2),

3 (1500 - y) + 5y = 5740

4500 -3y + 5y = 5740

-3y + 5y = 5740 - 4500

2y = 1240

$y = \frac{1240}{2}$

y = 620 adults

From 1 ,

x = 1500 - 620

x =880

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Which describes an isometric transformation? dilation/ rotation/ side length change/ angle change

Westkost [7]

Answer:

rotation

Step-by-step explanation:

"Isometric" means all measures stay the same.

dilation changes all lengths
<em>rotation changes no measures -- an isometric transformation</em>
side length change obviously changes the length of a side
angle change obviously changes the measure of an angle

7 0

3 years ago

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,

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

11 months ago

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm

sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

$\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}$