Change 4/50 into percent.

Mega middle school had a valentine dance four fifths of the students plus 7 teachers were there if there was 495 people at the d

maria [59]

495-7=488
488/4=122
122*5=610

The answer is 610 students attend the school.

6 0

3 years ago

Simplify the following expression as a monomial X^2y÷yx^2

Novay_Z [31]

$\bf ~\hspace{7em}\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
~\hspace{4.5em}
a^n\implies \cfrac{1}{a^{-n}}
~\hspace{4.5em}
\cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
x^2\div yx^2\implies \cfrac{x^2}{yx^2}\implies \cfrac{x^2\cdot x^{-2}}{y}\implies \cfrac{x^{2-2}}{y}\implies \cfrac{x^0}{y}\implies \cfrac{1}{y}$

8 0

2 years ago

Mr. Itol's assistant takes twice as long to complete a computer task as Mr. Itol. If it takes both experts 6 hours to complete t

Lesechka [4]

So, in order to solve this problem, I will set up an equation. First, let's make x= the # of hours it takes Mr. Itol to complete a computer task, and since his assistant takes twice as long, lets make 2x=the # of hours it takes the assistant to complete the task. We also know that it takes the two of them 6 hours to complete the task together. We can now set up the following equation:

$\frac{1}{2x} + \frac{1}{x} = \frac{1}{6}$

Now, in order to be able to solve this, we have to find the least common denominator, and then we can cancel out the denominators to solve for the top part. The LCD=6x and the new equation becomes:

$\frac{3}{6x} + \frac{6}{6x} = \frac{x}{6x}$

Now, we can get rid of the denominators, and we get:

$3+6=x$

So, x= 9

Now we know that Mr. Itol would take 9 hours to complete the task if he were to do it himself. To find how long it would take his assistant to complete the same task by himself, we just need to plug in the x-value into 2x:

$2(9)=18$

Therefore, your answer is that Mr. Itol would take 9 hours to do the job alone, while his assistant will take 18 hours (double the time) to do the job alone.

Hope this helps :)

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=4-3%283%29%5E%7B2%7D%20%5C%5C%20%5Csqrt%7B1%2B5" id="TexFormula1" title="4-3(3)^{2} \\ \sqrt{1

yarga [219]

This would be.
-23√6 all over 6.
Or: around -9.39

5 0

2 years ago

Find integrate (1/(xsqrt(1+(lnx)^2)),1,e,x) ...?

prohojiy [21]

Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as

(x² + x + 3 + x)/(x² + x + 3) = 1 + x/(x² + x + 3). 

Next, complete the square. 

x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2)² + (√(11)/2)² 

Also, for the x in the numerator 

x = x + 1/2 - 1/2. 

So 

(x² + 2x + 3)/(x² + x + 3) = 1 + (x + 1/2)/[(x + 1/2)² + (√(11)/2)²] - 1/2/[(x + 1/2)² + (√(11)/2)²]. 

Integrate term by term to get 

∫ (x² + 2x + 3)/(x² + x + 3) dx = x + (1/2) ln(x² + x + 3) - (1/√(11)) arctan(2(x + 1/2)/√(11)) + C 

b) Use the fact that ln(x) = 2 ln√(x). Then put u = √(x), du = 1/[2√(x)] dx. 

∫ ln(x)/√(x) dx = 4 ∫ ln u du = 4 u ln(u) - u + C = 4√(x) ln√(x) - √(x) + C 

= 2 √(x) ln(x) - √(x) + C. 

c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x. 

∫ 1/(e^(-x) + e^x) dx = ∫ e^x/(1 + e^(2x)) dx = ∫ du/(1 + u²) = arctan(u) + C 

= arctan(e^x) + C.

4 0

3 years ago