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3 years ago

What is 2\20 divided by 2

1 answer:

3 0

$\frac{2}{20} \div 2 \\ \\ \frac{1}{10} \div 2 \\ \\ \frac{1}{10} \times \frac{1}{2} \\ \\ \frac{1 \times 1}{10 \times 2} \\ \\ \frac{1}{20} \\ \\$

The final result is: 1/20 or 0.05.

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4+2g+5g=-66

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2 years ago

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Need help asap important! thanks so much!

a_sh-v [17]

Answer:

You would have $343.37 at the end of the 2 years.

Step-by-step explanation:

Interest earned is like bonus money the bank pays you just for keeping money

$\mathrm{Compund\:Interest\:Formula}:\quad A = P { \left( 1+ \dfrac{ r }{ n } \right) }^{ nt }$

P: the starting balance of the account (also called initial deposit, or principal)

A: the new balance in the account after N years.

t: the number of years or time

r: the interest rate, (in decimal form)

n: the number of times the interest is compounded each year.

Annually = each year = 1

P =$300, r = 7%, t = 2, n = 1, A = ?

Substitute the numbers into the "Compound Interest Formula".

$A = 300 { \left( 1+ \dfrac{ 0.07 }{ 1 } \right) }^{ 1 \times 2 }$

$\mathrm{Anything\:divided\:by\:1\:gives\:itself}$

$A = 300 { \left( 1+0.07 \right) }^{ 1 \times 2 }$

$\mathrm{Add\:1\:and\:0.07\:to\:get\:1.07}$

$A = 300 \times { 1.07 }^{ 1 \times 2 }$

$\mathrm{Multiply\:1\:and\:2\:to\:get\:2}$

$A = 300 \times { 1.07 }^{ 2 }$

$\mathrm{Calculate\:1.07\:to\:the\:power\:of\:2\:and\:get\:1.1449}$

$A = 300 \times 1.1449$

$\mathrm{Multiply\:300\:and\:1.1449\:to\:get\:343.47}$

$A = 343.47$

So you would have $343.37 at the end of the 2 years.

Look at the chart

$\mathrm{Year}\quad \mathrm{Year\: Interest}\quad \mathrm{Total\:Interest}\quad \mathrm{Balance}\\\quad1\quad\quad\quad $21.00\quad\quad\quad\quad $21.00\quad\quad\quad\quad $321.00\\\quad2\quad\quad\quad $22.47\quad\quad\quad\quad $43.47\quad\quad\quad\quad $343.47$

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2 years ago

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3 years ago

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3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi

stealth61 [152]

Applying our power rule gets us our first derivative,

$\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}$

simplifying a little bit,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

looking for critical points,

$\rm 0=2x^{-2/3}+4x^{1/3}$

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

$0=2x^{-2/3}\left(1+2x\right)$

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

$\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)$

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

and take our second derivative.

$\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Looking for inflection points,

$\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Again, pulling out the smaller power of x, and fractional part,

$\rm 0=-\frac43x^{-5/3}\left(1-x\right)$

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.

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3 years ago