Ask question

Ask question

All categories

2 years ago

6

Hayden has $20 to spend at the county fair. Admission to the fair is $5.00 per person, and tickets for food and games are $1.25

per ticket. What information is needed to calculate the amount of money that Hayden will have left after attending the fair?

1 answer:

Kamila [148]2 years ago

6 0

Answer:

prices of food and games and tickets

You might be interested in

ABC is an isosceles<br> AB= 3x-4 and BC= 5x-10<br><br> what is AB?

kakasveta [241]

3x-4=5x-10
Subtract 3x from both sides
-4=2x-10
Add 10 to both sides
6= 2x
X= 3
3 times 3-4
AB=5

3 0

3 years ago

Read 2 more answers

Six times a number increased by 5

sergey [27]

I'm not quite sure what you asking here but if your asking how to answer it in "math terms" than I'll be happy to explain :)

So "times" means multiply and "increase" means add so righting this in math terms would be
6x + 5
because it's six times "a number" which you can replace with x!

Let me know if you have any questions :3

7 0

3 years ago

Read 2 more answers

-7x + 13<br> HELP I WILL CHOOSE BRAINLIEST TO BEST ANSWER

Leni [432]

search it up than you will get the correct answer

3 0

3 years ago

Read 2 more answers

What is the number line of 2.5 + 4.5?

Harrizon [31]

Answer:

A

Step-by-step explanation:

It shows 2.5 and then it adds 4.5 to it

3 0

3 years ago

Read 2 more answers

Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)

baherus [9]

Step 1

Given;

$4x^2-y^2+24x+4y+28=0$

Required; To find the center that eliminates the linear terms

Step 2

$\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}$

Step 3

Substitute a,d,e into the vertex form

$\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}$ $\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}$

Step 4

Completing the square for -y²+4y

$\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}$

Step 5

Substitute a,d,e into the vertex form

$\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}$

Step 6

$\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}$

Step 7

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}$

Hence the answer is (-3,2)

4 0

1 year ago