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3 years ago

If you continue adding fractions according to this pattern when will you reach a sum of 2?

Mathematics

1 answer:

mr Goodwill [35]3 years ago

6 0

Answer:

You will never be able to reach the sum of 2

Step-by-step explanation:

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Can someone help with this question

Zielflug [23.3K]

Answer:

<1= 60

<2=120

<3=60

<4=60

<5=120

<6=120

<7=60

Step-by-step explanation:

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3 years ago

Simprury 20 s) 2 6 600J

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Answer:

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3 years ago

Determine whether the given value is a solution of the equation. 7w = 73; w = 10

kipiarov [429]

Step-by-step explanation:

When w=10, 7w=7×10=70

But the given value of 7w is 73.

So, the given value is not a solution of the equation.

8 0

3 years ago

plz help me and I kindly ask that u show work if possible? Plz don't answer if u don't know how to do the problem. PLZ help

soldier1979 [14.2K]

part A is just simple multiplication
since there are 2 6 sided number cubes there are 36 possible outcomes(6 sides *6 sides)

Part B is a little trickier. First off, you must find how to get seven.
1. 1+6 and 6+1

2. 2+5 and 5+2

3.3+4 and 4+3

therefore there are six combinations

you put that number over the total combos and get 6/36 or 1/6.

Hope this helps!

6 0

3 years ago

Read 2 more answers

The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.

tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that perpendicular lines have negative reciprocal slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

$\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}$

so, it passes through the midpoint of AB,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)$

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$

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3 years ago