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2 years ago

Ms. Hart paid $300 for 6 admission tickets and a $15 parking pass for an amusement park.

Mathematics

1 answer:

LUCKY_DIMON [66]2 years ago

7 0

Answer:

The cost of one admission ticket is <u>4</u><u>7</u><u>.</u><u>5</u>

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If I buy a stereo for $245 and this includes GST of 10%, what is the pre-GST price?

Ivan

Answer: $222.73

===========================================

Work Shown:

x = pre-GST price

10% of x = 0.10x = tax amount

x + 0.10x = 1.10x = post-GST price = 245

1.10x = 245

x = 245/1.10

x = 222.7272 approximately

x = 222.73 is the price before tax.

------------

Check:

10% of 222.73 = 0.10*222.73 = 22.273 = 22.27

The tax amount ($22.27) is added to the pre-GST price to get

22.27+222.73 = 245

which matches the post-GST price mentioned.

The answer is confirmed.

Or another way to confirm the answer is to calculate this

1.10*222.73 = 245.003 = 245

6 0

2 years ago

Read 2 more answers

Veronica wants to buy a house that costs $249,900. She is going to make a 20% down payment. How

rosijanka [135]

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▹ Answer

▹ Step-by-Step Explanation

$\frac{20}{100} * $249,900 \\\\= $49,980$

Hope this helps!

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6 0

2 years ago

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If Joe drives 186.83 miles on a business trip, and the reimbursement from his company is $13.08. At what rate is Joe's employer

kirill115 [55]

Answer:

7 cents/mile

Step-by-step explanation:

You are looking for a unit rate of cents per mile.

Change the dollar amount to cents, and divide by the number of miles.

$13.08 * (100 cents)/$ = 1308 cents

(1308 cents)/(183 miles) = 7.001 cents/mile

8 0

3 years ago

A cube has a side length of 12 m what is the volume of the cube?

AfilCa [17]

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➷ volume = length x width x height

volume = 12 x 12 x 12

volume = 1728

The volume is 1728 cm^3

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

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7 0

3 years ago

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Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

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3 years ago