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Andru [333]
3 years ago
6

What is three and one third divided by two aand one third?

Mathematics
1 answer:
grin007 [14]3 years ago
4 0
1 3/7 ~ One and three sevenths
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Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $35 and same-day tickets $30 c
Marina CMI [18]

Answer:

The number of same-day tickets sold is 15

The number of advance ticket sold is 20

Step-by-step explanation:

Given as :

The cost of each Advance ticket = $35

The cost of each same-day ticket = $30

The total cost of the tickets sold = $1150

The total number of tickets sold = 35

Let The number of advance ticket sold = a

And The number of same-day ticket sold = s

Now, According to question

The total number of tickets sold =The number of advance ticket sold + The number of same-day ticket sold

I.e a + s = 35                   .....A

The total number of tickets sold = The cost of each Advance ticket × The number of advance ticket sold + The cost of each same-day ticket × The number of same-day ticket sold

I.e $35 a + $30 s = $1150

or, 35 a + 30 s = 1150               .......B

Now, solving eq A and B

35 × ( a + s) - ( 35 a + 30 s ) = 35 × 35 - 1150

or, ( 35 a - 35 a ) + ( 35 s - 30 s ) = 1225 - 1150

or, 0 + 5 s = 75

∴  s = \frac{75}{5}

I.e s = 15

So, The number of same-day tickets sold = s = 15

Put the value of s in Eq A

So, a + 15 = 35

∴  a = 35 - 15

I.e a = 20

So, The number of advance ticket sold = a = 20

Hence The number of same-day tickets sold is 15

And  The number of advance ticket sold is 20   Answer

7 0
3 years ago
Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

8 0
3 years ago
Read 2 more answers
PLEASE HELP If You Are not SURE please don't answer PLEASE show work and steps
Ivanshal [37]
63+12x<15
12x<-48 (subtract 63 from both sides)
x<-4 (divide out the 12)
8 0
3 years ago
Read 2 more answers
Help
Rudiy27

Answer:

520

Step-by-step explanation:

must be the answer..

7 0
3 years ago
Read 2 more answers
ILL MAKE U BRAINLIEST
-BARSIC- [3]
The third box plot is your answer
5 0
3 years ago
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