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torisob [31]
3 years ago
6

How do I get the answer to this problem

Mathematics
2 answers:
Mashcka [7]3 years ago
5 0

Answer:\frac{19}{20} - \frac{10}{20} - \frac{2}{20}

\frac{7}{20}

Step-by-step explanation:

Since the Least Common Denominator is 20 (or the biggest number on the bottom in your case), you can multiply 4 and 10 by their corresponding number to get 20.

4 * 5 = 20

10 * 2 = 20

Now multiply the top and bottom by 5 and 2

\frac{5}{5} *\frac{2}{4} = \frac{10}{20}[tex]\frac{2}{2} *\frac{1}{10} = \frac{2}{20}Now subtract everything.[tex]\frac{19}{20} - \frac{10}{20} - \frac{2}{20}

Its equals \frac{7}{20} .

N76 [4]3 years ago
3 0

Answer:

You do it, or ask someone

Step-by-step explanation:

Because. Just because. That's logic.

jk.... 19/20 - 10/20 - 2/20

       19/20 - 12/20

       7/20

     

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Alona [7]
She made 200 dollars on commission. 2,500*1.08 = 2,700

I hope this helps :)
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3 years ago
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What are two other equivalent forms of 3(x-1)(x-2)
katen-ka-za [31]

You can multiply it out to get

... 3x² - 9x + 6

and you can write it in vertex form as

... 3(x -3/2)² -3/4

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4 years ago
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EastWind [94]

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Step-by-step explanation:

This will shift the graph 5 units right

8 0
2 years ago
Marco is studying a type of mold that grows at a fast rate. He created the function f(x) = 345(1.30)x to model the number of mol
Ivahew [28]

1.30 represents the rate of growth of the mold spores. This decimal is a representation of the percentage growth, which would be 130%.


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345(1.30)^4 = \boxed{985.35}


Rounded to the nearest ones value, there will be 985 mold spores after 4 weeks.

8 0
3 years ago
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Solve the given initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1
Yuri [45]
Let's check if the ODE is exact. To do that, we want to show that if

\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0

then M_y=N_x. We have

M_y=2(x+y)
N_x=2y+2x=2(x+y)

so the equation is indeed exact. We're looking for a solution of the form \Psi(x,y)=C. Computing the total differential yields the original ODE,

\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0
\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}

Integrate both sides of the first PDE with respect to x; then

\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)

where f(y) is a function of y alone. Differentiate this with respect to y so that

\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)
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f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C

So the solution to this ODE is

\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C

i.e.


\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C
6 0
3 years ago
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