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Natalija [7]
3 years ago
8

Eight people sharing 7 candy bars how many other faction would they get

Mathematics
2 answers:
OLEGan [10]3 years ago
8 0
If you mean how much would they get in fraction form, everyone will get 7/8 of a candy bar
bonufazy [111]3 years ago
5 0
For each person they will get 7/56 of a candy bar
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What is 28% of 50? <br><br> SHOW YOUR WORK PLEASE!
miv72 [106K]

Answer:

14

Step-by-step explanation:

.28 times 50 equals 14

7 0
3 years ago
Please help with A.
Furkat [3]

Answer:

ok so I dont have the answer but I suggest if u can u do the tutor they help a lot u can only ask 1 question per session but trust me it's helpful

they also say the answer at the end

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3 0
3 years ago
Can someone check whether its correct or no? this is supposed to be the steps in integration by parts​
Gwar [14]

Answer:

\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

Step-by-step explanation:

\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}

Given integral:

\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x

\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:

\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x

Using <u>integration by parts</u>:

\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

Therefore:

\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}

\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:

\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}

Therefore:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x

\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:

\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}

Divide both sides by 2:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}

Rewrite in the same format as the given integral:

\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

5 0
2 years ago
The quantity demanded x of a certain brand of DVD player is 3000/week when the unit price p is $485. For each decrease in unit p
ANTONII [103]

Answer:

Demand Equation: q = 15125 - 25p.

Supply Equation: 80p - 24000 = 9q.

Equilibrium Price: $525.

Equilibrium Quantity: 2000 units.

Step-by-step explanation:

To solve this question, first, the demand equation has to be calculated. Let price be on the y-axis and quantity by on the x-axis. It is given that p = $485 when q = 3000 units. This can also be written as (q₁, p₁) = (3000, 485). It is also given that when the price goes down by $20, the quantity increases by 500 units. Therefore, (q₂, p₂) = (3500, 465). Due to the statement "For each decrease in unit price of $20 below $485, the quantity demanded increases by 500 units", the demand function will be linear i.e. a straight line. This is because the word "for each" has been used. Therefore, finding the equation of the demand function:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 485)/(q - 3000) = (465 - 485)/(3500 - 3000).

(p - 485)/(q - 3000) = (-20)/(500).

(p - 485)/(q - 3000) = -1/25.

25*(p - 485) = -1*(q - 3000).

25p - 12125 = -q + 3000.

25p + q = 15125.

q = 15125 - 25p. (Demand Equation).

Second, find the supply equation. It is given that the suppliers will not sell the product below or at the price of $300. Therefore, (q₁, p₁) = (0, 300). Also, the suppliers will sell 2000 units if the price is $525. Therefore, (q₂, p₂) = (2000, 525). Since it is mentioned that supply equation is linear, therefore:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 300)/(q - 0) = (525 - 300)/(2000 - 0).

(p - 300)/(q) = (225)/(2000).

(p - 300)/(q) = 9/80.

80*(p - 300) = 9*(q).

80p - 24000 = 9q. (Supply Equation).

Equilibrium exists when demand = supply. This means that the demand and the supply equations have to be solved simultaneously. Therefore, put demand equation in supply equation:

80p - 24000 = 9(15125 - 25p).

80p - 24000 = 136125 - 225p.

305p = 160125.

p = $525.

Put p = $525 in demand equation:

q = 15125 - 25p.

q = 15125 - 25(525).

q = 15125 - 13125.

q = 2000 units.

To summarize:

Demand Equation: q = 15125 - 25p!!!

Supply Equation: 80p - 24000 = 9q!!!

Equilibrium Price: $525!!!

Equilibrium Quantity: 2000 units!!!

8 0
3 years ago
Translate the following points 5 units up. <br> (-5, -2) (-2, 1) (-4, -7)
makkiz [27]
(-5, 3) (-2, 6) (-4, -2)
those will be your answers
8 0
3 years ago
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