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Kaylis [27]
3 years ago
8

Akira buys 10 tickets to a show. She also pays a $5 parking fee. If she spent $35 in total, how much did each ticket cost? Use x

to represent the cost of one ticket.
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
7 0

Answer: Each ticket cost 3 dollars

Step-by-step explanation: First take 35 and subtract 5 which equals 30

then divide 30 by 10 and you get 3

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Find the midpoint of AB for: A(7,0)|and B(0,3)​
liberstina [14]

Answer:

\boxed {\boxed {\sf (\frac {7}{2}, \frac{3}{2}) \ or \ (3,5, 1.5) }}

Step-by-step explanation:

The midpoint is essentially a point with the average of the 2 x-coordinates and the 2 y-coordinates.

The formula is:

(\frac {x_1+x_2}{2}, \frac{y_1+y_2}{2})

We are given two points: A (7,0) and B (0, 3). Remember points are written as (x, y).

Therefore,

x_1= 7 \\y_1=0 \\x_2=0 \\x_2=3

Substitute the values into the formula.

(\frac {7+0}{2}, \frac{0+3}{2})

Solve the numerators first.

(\frac {7}{2}, \frac{3}{2})

The midpoint can be left like this because the fractions are reduced, but it can be written as decimals too.

(3.5, 1.5)

3 0
3 years ago
A new car is sold with or without antilock brakes, with or without a CD player, with or without electric seats, and with or with
True [87]

Answer:

8 different choices

Step-by-step explanation:

With or without antibrake=2 ways

With or without player=2ways

With or without sunroof=2 ways

Total number of ways=2×2×2=8 ways

3 0
3 years ago
Find a normal subgroup of v which is not normal in a4.
Tpy6a [65]

he elements of the Klein <span>44</span>-group sitting inside <span><span>A4</span><span>A4</span></span> are precisely the identity, and all elements of <span><span>A4</span><span>A4</span></span>of the form <span><span>(ij)(kℓ)</span><span>(ij)(kℓ)</span></span> (the product of two disjoint transpositions).

Since conjugation in <span><span>Sn</span><span>Sn</span></span> (and therefore in <span><span>An</span><span>An</span></span>) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.

6 0
3 years ago
How do you simplify this problem
vampirchik [111]
Reduce the fraction with 2
\frac{2 \times x {}^{ - 2} y {}^{3}z {}^{ - 1}  }{xy}  \\  \\  \frac{3y {}^{2}z {}^{ - 1}  }{x {}^{3} }  \\  \\  \frac{3y {}^{2} }{x {}^{3}   z}
7 0
3 years ago
Am I correct? Please correct me if I'm wrong
laiz [17]

Answer:

Great job

Step-by-step explanation:

3 0
3 years ago
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