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AVprozaik [17]
2 years ago
10

Situation: Tickets to a baseball game are $20 for an adult and $15 for a student. On Saturdays there is a 3 dollar discount on t

ickets. Create 3 problems and solve 1 of them.
Mathematics
2 answers:
polet [3.4K]2 years ago
6 0
Problem 1. Three students and one adult went to the baseball game on a Saturday, how much was their total bill?

problem 2. Five students and five adults went to the baseball game on a Monday, how much was their total bill?

problem 3: One student and two adults went to the baseball game on a Friday, how much was their total bill?

answer: their total bill was $55

explanation: two adults ($20 + $20 = $40) + one student ($40 + $15 = $55) the discount does not apply because they did not attend the baseball game on a Saturday
sveticcg [70]2 years ago
5 0
<h3>                                     Answer: </h3>

                                                ↓

       Problem 1. Three students and one adult went to the baseball game on a Saturday, how much was their total bill?

Problem 2. Five students and five adults went to the baseball game on a Monday, how much was their total bill?

Problem 3: One student and two adults went to the baseball game on a Friday, how much was their total bill?

Answer: their total bill was $55

<h3>                               Explanation: </h3>

                                              ↓

two adults ($20 + $20 = $40) + one student ($40 + $15 = $55) the discount does not apply because they did not attend the baseball game on a Saturday.

<h2>              Can you brainliest me?</h2>

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Solve the following multiplication and division problems.
algol13

Answer:

a. 22.5 km

b. 267.52 m

c. 95 m

d. (4.2)(10^{-5}) km

Step-by-step explanation:

a. You can convert from hectometers to kilometers, and then you can make the multiplication:

\frac{5hm}{10}=0.5km

(3km)(0.5km)(15)=22.5km

b. You can convert from centimeters to meters, and then you can make the multiplication:

\frac{44cm}{100}=0.44m\\(19m)(0.44m)(32)=267.52m

c. You only need to divide 1,140 meters by 12:

\frac{1,140m}{12})=95m

d. You can convert from hectometers to kilometers, from decameters to kilometers and from meters to kilometers, and divide this by 15:

\frac{6hm}{10}=0.6km

\frac{7dam}{100}=0.07km

\frac{5m}{1000}=0.005km

\frac{(3km)(0.6km)(0.07km)(0.005km)}{15}=(4.2)(10^{-5}) km

5 0
3 years ago
The shape below is a horizontal cross section of a figure. Select all the possible figures that could have this shape as a cross
Dvinal [7]

Answer:

D. Rectangular prism

Step-by-step explanation:

I believe the answer is D. Rectangular prism

3 0
3 years ago
Find the length of the hypotenuse of an isosceles right triangle whose legs are 1 unit in length.
stiks02 [169]

check the picture below

and isosceles triangle, has two equal sides, thus, sides "a" and "b" for the triangle to be an isosceles, have to be equal

6 0
3 years ago
Read 2 more answers
Using Ratios to Solve Problems - Item 1457 Question 4 of 7 Evan's family drove to a theme park for vacation. They drove the same
azamat

Answer:

150 miles

distance travelled on the third day = 150 miles

Step-by-step explanation:

Note: it was given that they drove the same speed throughout the trip. That means their speed is constant for all days;

For the first day

distance = 300 miles

time = 6 hours

Speed = distance/time

Speed = 300/6 = 50 mph

on the third day;

time = 3 hours

Speed = speed on first day = 50 mph

distance = speed × time

distance = 50 mph × 3 hours

distance = 150 miles

distance travelled on the third day = 150 miles

5 0
3 years ago
A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.
olga55 [171]
We know that:

(x-a)^2+(y-b)^2=r^2

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}

and all we have to do is solve it for a, b and r.

There will be:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\&#10;\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\&#10;\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\&#10;

From equations (II) and (III) we have:

\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}

and from (I) and (II):

\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}

Now we can easly calculate a and b:

\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}

Finally we calculate r^2:

a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}

And the equation of the circle is:

(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}
7 0
3 years ago
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