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GREYUIT [131]
2 years ago
6

Pam went to the fair. She went on the same 6 times and used the same number of tickets each time. She used 18 tickets. How many

tickets did she use each time she went on the ride?
Mathematics
1 answer:
Gekata [30.6K]2 years ago
6 0

Answer:

3 tickets

Step-by-step explanation:

If you did 18/6 it would cost 3 tickets each time on the ride.

18÷6=3 tickets

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Which of the following are exterior angles? Check all that apply.
Marina86 [1]

Answer:

A

C

D

E

Step-by-step explanation:

Exterior angles can be described as the angles that are formed between the side of a polygon and the extended adjacent side of the polygon.

Or an exterior angle is the angle that is not inside the triangle formed.

The angles inside the triangle are interior angles.

Exterior angles are :

2

3

4

6

Interior angles are :

1

5

6 0
2 years ago
HELP ME ASAP! Will give BRAINLIEST!
Nataliya [291]

Answer:

Linear

Step-by-step explanation:

It is just a never ending straight line. no complications in that.

6 0
3 years ago
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Calculate p² +3q when p=2 and q=2​
QveST [7]

Answer:

10

Step-by-step explanation:

p² +3q

= 2² +3*2

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4 0
3 years ago
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Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?
cluponka [151]

Under the given transformation, the Jacobian and its determinant are

\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2

so that

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv

where D' is the region D transformed into the u-v plane. The remaining integral is the twice the area of D'.

Now, the integral over D is

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y

but through the given transformation, the boundary of D' is the set of equations,

\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}

We require that u>0, and the last equation tells us that we would also need v>0. This means 1\le v\le\sqrt2 and 0, so that the integral over D' is

\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6

4 0
2 years ago
1627187 to nearest ten
Orlov [11]
1627190 there's your answer.
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2 years ago
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