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2 years ago

9

I need this homework

1 answer:

galina1969 [7]2 years ago

6 0

Ummmmm... what is the question can u plz let me know I can't see the full question

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A music website charges x dollars for individual songs and y dollars for an entire album. Person A pays $25.92 to download 6 ind

EleoNora [17]

Amount charged for downloading individual songs = x dollars
Amount charged for downloading an entire song album = y dollars
Person A:
6x + 2y = 25.92
3x + y = 12.96
y = 12.96 - 3x
Person B:
4x + 3y = 33.93
Putting the value of y from the first equation in the second we get
x = 0.99 dollars
Putting the value of x in the first equation we get
y = 12.96 - 3x
= 12.96 - 2.97
= 9.99 dollars

7 0

2 years ago

Please help me wiht this

Alex17521 [72]

Answer:

Kevin

Step-by-step explanation:

5 0

3 years ago

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:

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

7 0

2 years ago

OK I'm really confused.

pickupchik [31]

Answer:

i am pretty sure the answer is UT.

Step-by-step explanation:

this is because all the letters have linear connections to the other letter, all except UT do.

i hope this helps, tell me if this is wrong.

8 0

2 years ago

If the circumference of a circle is 37.7 what is the diameter

Arada [10]

The diameter is 12 :)

6 0

2 years ago