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8_murik_8 [283]
3 years ago
14

Edmund printed 36 photos at a photo kiosk. he paid a total of $ 10.08. what was the cost per photo?

Mathematics
2 answers:
Oksanka [162]3 years ago
7 0
Each photo cost 28 cents.
Schach [20]3 years ago
7 0
10.08 = 1008 cents

1008 cents = 36 photos

Divide both sides by 36 to find out the cost for 1 photo

28 cents = 1 photo

The cost per photo was 28 cents. 
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Anyone know how to do this ????
SOVA2 [1]

There you go. Hope this helps.

5 0
2 years ago
Please help and explain how to do this
Mnenie [13.5K]

Answer:

  4√3

Step-by-step explanation:

The distance formula applies. It tells you ...

  distance = √((x2 -x1)² +(y2 -y1)²)

Filling in the given values, you have ...

  distance = √((-√32 -(-4√2))² +(2√3 -(-√12))²)

  = √((-4√2+4√2)² +(2√3 +2√3)²)

  = √(0 + (4√3)²)

  distance = 4√3

___

We make use of the fact that ...

  a\sqrt{b}=\sqrt{a^2b}\\\\\sqrt{32}=\sqrt{16\cdot 2}=4\sqrt{2}\\\\2\sqrt{3}=\sqrt{2^2\cdot 3}=\sqrt{12}

3 0
3 years ago
S
Novay_Z [31]

Step-by-step explanation:

s is inversely proportional to t

If s= 0.6 , t= 4

s=k/t

0.6= k/4

k=2.4

If s=12, then

t=k/s

t=2.4/12

t=0.05

8 0
2 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
3 years ago
Someone help me with this question!
Oksi-84 [34.3K]
The “1/2” caused the parabola to be vertically compressed meaning it got wider.

and the “-2” caused the parabola to move 2 units to the right

Therefore the correct answer should be C.
8 0
3 years ago
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