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OverLord2011 [107]
3 years ago
14

Jose paid $47.00 for 4 movie tickets. Each ticket cost the same amount. What was the cost of each movie ticket in dollars and ce

nts?
Mathematics
2 answers:
scoundrel [369]3 years ago
4 0
Each ticket costed 11.75$
Stells [14]3 years ago
4 0
$11.75

If you need more information, just ask. I'll be happy to help. Also, if this helped you can you please say thanks + rate and put as brainliest (2 people have to answer first), trying to get a new badge!! Thank you so much!

Hope this helped! Have a great day!
You might be interested in
9 3/10 - 5 3/4 as a mixed number in simplest form
Artemon [7]

Answer:

3 1/20

Step-by-step explanation:

First, we will convert the mixed numbers into improper fractions...

9 3/10 (9*10=90+3=93) so the improper fraction is 93/10

5 3/4 (5*4=20+3=23) therefore the improper fraction is 23/4

We then want to make the denominator the same...

93/10 (*4) = 372/40

23/4 (*10) = 230/40

We then put it into the question...

372/40 - 230/40 = 142/40

Lastly, we now need to turn this into a mixed number...

142/40=3.55 so the answer is 3 2/40

3 2/40 in it's simplest form is 3 1/20

Hope this helps!

6 0
3 years ago
Read 2 more answers
PLEASE HELP ME!!! QUICKLY!!
podryga [215]

Answer:

4

Step-by-step explanation:

2L=w

32=LxW

so you put the 2L as the w

which is 32=Lx2L

32=2L^2

16=L^2

4=L

*To find the width you input 4 as L*

So 8=W

The shortest side would be 4

3 0
2 years ago
The book fair is this week, and there's a special deal for teachers. For every 5 hardcover
zhuklara [117]

Answer:14

Step-by-step explanation:

6 0
3 years ago
Ming has two job offers, in different states. The first job (a marketing manager) earns a salary of $51,000 and there is no stat
elixir [45]

Answer:

A. 1

Step-by-step explanation:

3 0
3 years ago
Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
Anit [1.1K]

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

4 0
3 years ago
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