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Hoochie [10]
3 years ago
15

It costs $70 for 50 tickets. How much does each ticket cost?

Mathematics
2 answers:
BigorU [14]3 years ago
6 0
If you do $70 divided by 50, you would get 1.4, and 1.4 times 50 would be $70. So, each ticket would cost $1.40.
liberstina [14]3 years ago
5 0
1 dollar and 40 cents
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A $65 coat is now on sale for $52. What percent discount is given?
leonid [27]
20%, because if 65 is 100, then 52 is 80% out of 65; so 100 - 80 is 20
3 0
3 years ago
Read 2 more answers
What is the value of x
My name is Ann [436]
X=3



Explanation:

A straight line equals 180°, and they give you the one angle of 56°, so you know 56+ the other equation has to equal 180

56+4x+112=180
168+4x=180

Subtract 168 from both sides

4x=12

Divide 4 from both sides

X=3


You can check by filling it in.


56+4*3+112=

12+56+112=180
5 0
3 years ago
15℅ of 6th grade teachers give homework over the holiday break. If 12 teachers give homework, how many total 6th grade teachers
hoa [83]

Answer:

12

Step-by-step explanation:

Its a trick question, 15% of 6th grade teachers give HW, but there are 12 teachers. its not gonna be 15% of 12, because you would also get a fraction of a teacher.

4 0
3 years ago
Hillary pours 10 cups of orange juice into glasses that hold 1 2/3 cup each. How many glasses does Hillary fill?
Svetach [21]

10 =  \frac{30}{3 }  \\ 1 +  \frac{2}{3}  =  \frac{5}{3}  \\ \frac{30}{3}  \div  \frac{5}{3}  =  \frac{30}{3}  \times  \frac{3}{5}  = 8glasses

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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) =  δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)]     since  [cos²(6θ) + sin²(6θ)] = 1

6 0
4 years ago
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