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2 years ago

A ticket company charges a 4% service fee on all orders. How much is the service fee for a ticket that costs $65?

Mathematics

1 answer:

solong [7]2 years ago

6 0

Answer:

$16.25

Step-by-step explanation:

4 goes into 65 16 r1

which = 16.25

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At the city museum, child admission is $5.20 and adult admission is $9.90 . On Monday, twice as many adult tickets as child tick

DiKsa [7]

To solve this, I would just find what two adult tickets and one child ticket wold be worth.

To find that, just do:

5.20 + 9.90 * 2, which makes 25.

So now you just find how many 25s are in 900, or 36.

So 36 children's tickets were sold that day.

Hope I helped :)

7 0

3 years ago

1+1<br> ......................................................................................

aleksley [76]

Answer:

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8 0

3 years ago

Read 2 more answers

A 140 kg camera is suspended by two wires over a 40 metre wide football field to get shots of the action from above. At one poin

Katen [24]

Answer:

Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

$T_1cos\theta_1-T_2cos\theta_2=0$ (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

$T_1sin\theta_1+T_2sin\theta_2-W=0$ (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

$T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2$

Then, you sum the equations:

$T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2$ (3)

Next, you use the following identity:

$sin^2\theta+cos^2\theta=1$

and you obtain in the equation (3):

$T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°$

With this values you can calculate the value of the another angle, by using the equation (1):

$\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°$

Now, you can calculate the length of each cable by using the information about the width of the football field. You use the following trigonometric relation:

$l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\$

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

3 0

2 years ago

What is the lenght of PQ<br> P is 2,1<br> Q is 5,6

Elodia [21]

Answer:

7,7

Step-by-step explanation:

Possibly 7,7 becuase p+q=pq, therefor 2,1+5,6=7,7

4 0

3 years ago

Plzzzzzzzzzzz help me

oksano4ka [1.4K]

The second table does not represent a function

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3 years ago