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xxTIMURxx [149]
2 years ago
7

Write a division expression with quotient that is greater than 8 divided 0.001

Mathematics
1 answer:
Jet001 [13]2 years ago
5 0

Answer:

Expression is '98 divided by 0.005' i.e. \frac{98}{0.005}.

Step-by-step explanation:

We have that,

The quotient of 8 divided by 0.001 is given by,

\frac{8}{0.001} = \frac{8\times 1000}{1} = 8000.

It is required to find a division expression having quotient greater than 8000.

Let us consider,

'98 divided by 0.005'

i.e. \frac{98}{0.005}

i.e. \frac{98\times 1000}{5}

i.e. 98 × 200

i.e. 19,600

Thus, the get the quotient of the new expression 19,600 > 8000.

Hence, the required expression is '98 divided by 0.005' i.e. \frac{98}{0.005}.

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Solve the application problem. Round to the nearest percent. An ice cream shop increased its number of flavors from 20 to 30. Wh
stiv31 [10]

Answer:

33%

Step-by-step explanation:

Why it is 33% is you minus 20 and 30 to get 10 then you just put that on top of 30 to get 10/30 and you will get 33%.

4 0
3 years ago
Select the correct answer which function is increasing at the highest rate?
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Answer:

D.since the gragh is linear

4 0
3 years ago
Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a
Crank

I'll assume the ODE is actually

y''+(x-2)y'+y=0

Look for a series solution centered at x=2, with

y=\displaystyle\sum_{n\ge0}c_n(x-2)^n

\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n

\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n

with c_0=y(2)=2 and c_1=y'(2)=0.

Substituting the series into the ODE gives

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0

\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}

  • If n=2k for integers k\ge0, then

k=0\implies n=0\implies c_0=c_0

k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}

k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}

k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}

and so on, with

c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}

  • If n=2k+1, we have c_{2k+1}=0 for all k\ge0 because c_1=0 causes every odd-indexed coefficient to vanish.

So we have

y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}

Recall that

e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}

The solution we found can then be written as

y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k

\implies\boxed{y(x)=2e^{-(x-2)^2/2}}

6 0
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vichka [17]

Answer:

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Step-by-step explanation:

=-9[piojkhlbjvh gcfxsae vndhgcnboiybsgdtjygncvnftu

3 0
3 years ago
A $4,500 savings account earns $540 in interest. What is the percent of increase on the original value? Show work please​
gavmur [86]

540:4500*100 =

(540*100):4500 =

54000:4500 = 12%

6 0
2 years ago
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