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Ghella [55]
4 years ago
6

Simplify -3x+1y+8x-8y

Mathematics
1 answer:
lana66690 [7]4 years ago
8 0
-3x+8x+1y-8y
5x-7y , is the answer
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Solve the differential equation dy dx equals the quotient of x squared and y squared for y = f(x) with the initial condition y(0
Gnoma [55]
\displaystyle\frac{dy}{dx} = \frac{x^2}{y^2}\ \Rightarrow\ y^2 dy = x^2 dx\ \Rightarrow\ \int y^2 dy = \int x^2 dx\ \Rightarrow\textstyle\ \frac{1}{3}y^3 = \frac{1}{3}x^3 + C. \\ \\
\text{Now } y(0) = 2\ \Rightarrow\ \frac{1}{3}(2)^3 = \frac{1}{3}(0)^3 + C\ \Rightarrow\ \frac{8}{3} = C,\text{ so } \frac{1}{3}y^3 = \frac{1}{3}x^3 + \frac{8}{3}. \\ \\
y^3 = x^3 + 8\ \Rightarrow\ y = \sqrt[3]{x^3 + 8}
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3 years ago
Find x, assume that any segment that appears to be tangent is tangent.​
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Simplify for this question !
PolarNik [594]

Answer:

=6x^9-3 [12/2=6]

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6 0
3 years ago
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Multiply by hundreds: 545 times 200
fiasKO [112]

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109000

Step-by-step explanation:

5 0
3 years ago
How much should a family deposit at the end of every 6 months in order to have $4000 at the end of 3 years? The account pays 5.9
ra1l [238]

We are asked to determine the present value of an annuity that is paid at the end of each period. Therefore, we need to use the formula for present value ordinary, which is:

PV_{ord}=C(\frac{1-(1+i)^{-kn}}{\frac{i}{k}})

Where:

\begin{gathered} C=\text{ payments each period} \\ i=\text{ interest rate} \\ n=\text{ number of periods} \\ k=\text{ number of times the interest is compounded} \end{gathered}

Since the interest is compounded semi-annually this means that it is compounded 2 times a year, therefore, k = 2. Now we need to convert the interest rate into decimal form. To do that we will divide the interest rate by 100:

\frac{5.9}{100}=0.059

Now we substitute the values:

PV_{ord}=4000(\frac{1-(1+0.059)^{-2(3)}}{\frac{0.059}{2}})

Now we solve the operations, we get:

PV_{\text{ord}}=39462.50

Therefore, the present value must be $39462.50

8 0
2 years ago
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