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3 years ago

which is the rationalized form of the expression the square root of x divided by the square root of x +the square root of 7

Mathematics

1 answer:

Ilya [14]3 years ago

8 0

Answer:

I assume your goal is to rationalize the denominator -

So you need to multiply the denominator by its conjugate.

So we have:

$\frac{\sqrt{x} }{\sqrt{x}-\sqrt{7} } *\frac{\sqrt{x} +\sqrt{7} }{\sqrt{x} +\sqrt{7}}$

$\frac{x-\sqrt{7}\sqrt{x} }{x-7}$

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A square has sides 5.3cm long. Work out the area of the square

mezya [45]

Answer:

Area of square = 28.09 square cm.

Step-by-step explanation:

Given : A square has sides 5.3 cm long.

To find : Work out the area of the square.

Solution : We have given

Side of square = 5 .3 cm .

Area of square = side * side .

Area of square = 5.3 * 5.3 .

Area of square = 28.09 square cm.

Therefore, Area of square = 28.09 square cm.

4 0

3 years ago

Julia has 3 ribbons that are 12 inches each. how many 4 inch sections can she cut from the ribbon.

Ber [7]

Total length of all ribbons = 12 * 3 = 36 inch

Then, number of ribbons = 36/4 = 9

In short, Your Answer would be: 9 sections

Hope this helps!

6 0

3 years ago

Is a D a failing grade?

s344n2d4d5 [400]

Answer:

Yes anything under a C

Step-by-step explanation:

6 0

3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Dr. Blumen invested $5,000 part of it was invested in bonds at a rate of 6% in the rest was invested in a money market at the ra

abruzzese [7]

So... let's say the amounts invested were "a" at 6% and "b" at 7.5%.

ok.. hmm what's 6% of a? well, (6/100) * a or 0.06a.
what's 7.5% of b? well, (7.5/100) * b or 0.075b.

now... we know, whatever "a" and "b" are, they total the investment of 5000 bucks, thus a + b = 5000

and the interest yielded was 337.50, thus 0.06a + 0.075b = 337.50

thus $\bf \begin{cases}
a+b=5000\implies \boxed{b}=5000-a\\
0.06a+0.075b=337.50\\
----------\\
0.06a+0.075\left( \boxed{5000-a} \right)=337.50
\end{cases}$

solve for "a", to see how much was invested at 6%.

what about "b"? well, b = 5000 - a.

6 0

3 years ago

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