All Answers 358698 Answers

$$\small{\text{ The fraction is
$\dfrac{79}{90}
$}}$$

jo has 12 stamps, 3 U.S stamps and 9 foreign stamps.

 There is a malfunction in the 25-450 computer subprocessors, but eventually they will answer.

4/5 x 1 1/6

Turn 1 1/6 into an improper fraction

4/5 x 7/6

Multiply across

28/30

Simplify

14/15

And there's your answer! :)

Infinity is a number greater than any assignable quantity or countable number

The angle of depression is congruent to the angle of elevation (this assumes the object is close enough to the observer so that the horizontals for the observer and the object are effectively parallel; this would not be the case for an astronaut in orbit around the earth observing an object on the ground)

The formula for the sum of an infinite series is:  Sum  =  a/(1 - r) provided that -1 < r <  1.

As r gets closer and closer to -1, the sum gets closer and closer to 4.5; so x can be 5 (if r = -4/5).

sorry, but what is the fraction for this?

$$\small{
\text{sum $ \ s = \underbrace{(7.9*10^{-1})}_{=a} + (7.9* 10^{-1} ) * 10^{-1} + (7.9* 10^{-1} ) * 10^{-2} + (7.9* 10^{-1} ) * 10^{-3} + \dots + (7.9* 10^{-1} ) * 10^{-(n-1)}
$
}}$\\\\\\$
\small{\text{
$
r = \dfrac{ (7.9* 10^{-1} ) * 10^{-1} }{ (7.9* 10^{-1} )} = 10^{-1}
$
}}$\\\\$
\small{\text{
$
s = \dfrac{ a }{ 1-r } = \dfrac{(7.9* 10^{-1} ) }{1-10^{-1} }
$
}}$\\\\$
\small{\text{
$
s = \dfrac{(7.9* 10^{-1} ) }{1- 10^{-1} } * \dfrac{ 10^{1}}{10^{1} }
$
}}$\\\\$
\small{\text{
$
s = \dfrac{ 7.9 } { 10^{1} -1} = \dfrac{7.9}{9}
$
}}$\\\\$
\small{\text{
$
s = \dfrac{7.9}{9} = 0.87777777777777777777777777\overline{7} = 0.8\overline{7}
$
}}$$

it is 19, but refering to the joke... 21 : )

19

If you add those together, you would get about 2.6. If I am correctly approaching this, then it would be closer to two because one is 1.6 away. Two is only 0.6 away. If I understood this question wrong, please comment back.

nothing,everything and most importantly somthing

The age of a man 3 years ago was 11 times the age of his son. The ratio of their ages after 15 years will be 2:1

And I assume you want to find their current ages??

Let x and y be the son's and father's ages, now...so we have

11(x-3)  = y-3 →   11x - 33 = y - 3  →   11x - 30 = y    (1)

2(x + 15) = y + 15 →  2x + 30 = y + 15 →  2x + 15 = y

And setting the "y's" equal, we have

11x - 30 = 2x + 15  rearrange

9x = 45   solve for x

x = 5, and that's the son's age now

And using (1)

11(5) - 30 = y  = 25  and that's the father's age now

The age of a man 3 years ago was 11 times the age of his son. The ratio of their ages after 15 years will be 2:1

$$\small{\text{ s = son and m = man }$\\\\$
\begin{array}{lrcllrcl}
(1) & (m-3) &=& 11*(s-3) & | \quad (m-3) &=& 11s-33\\
(2) & \frac{m+15}{s+15} &=& \frac{2}{1} & | \quad m+15 &=& 2s+30\\
\hline
(2)-(1):& 18 &=& -9s+63 & | :9\\
&2 &=& -s+7\\
& s&=&5 \\
\hline
(1): & m-3 &=& 55-33\\
& m &=& 3+22\\
& m &=& 25 \\
\hline
\end{array}$$

The age of the man is 25 and the age of his son is 5.

NEVER USE WIKIPEDIA

 $$\small{\text{Arithmetical sequence : } }\boxed{\ a_n=a_1+(n-1)*d\ } \\\\
a_1=7 \text{ and } d = 4 \text{ , so we have } a_n = 7 + (n-1) * 4 = 3+4n \\
7+0*4,\ 7+1*4,\ 7+2*4,\ 7+3*4, \ +\dots+,\ 7+n*4$$

Neither!

This is simple interest because the money is actually PAID each day.  Each dày the money is growing so if you look at what you have at the end compared to what  you started with it will behave like compound interest.

You are not doing it wrong exacly but you are not doing it 182 times either. That's how many times simple interest has been earned.

The only sensible way to do this is with the compound interest formula.

In order to go from radians to degrees you have to multiply the value by 180/∏. 

2 Many

Sasini meant to type "width" rather than "weight'.....