All Answers 358580 Answers

If "f" denotes the fraction of save bonds,

Then: 17.2 = 0.07 * (f*190) + 0.1 * (1-f) * 190

So f = (0.1 - 17.2 / 190) / 0.03

Therefore: about 31.5% of save bounds and 68.5% of the other bond.

 x-y+2z = -1   (1)

     2x+z = 0   →  z = -2x    (2)

 x+5y+z = 5     (3)

Putting (2)  into (1) and (3), we have

x - y + 2(-2x)  = -1

x +  5y - 2x  = 5         simplify  both of these

-3x - y   = -1    →   y = -3x + 1    (4)

-x  + 5y   = 5      (5)

Putting  (4)  into (5), we have

-x + 5(-3x + 1)  = 5

-16x + 5  =  5      subtract 5 from both sides

-16x = 0       → x  =  0

And using (4)

y = -3(0)  + 1    = 1

And using (2)

z = -2(0)   = 0 

So   {x, y, z}   =  {0, 1, 0}

x-y+2z=-1

y=x+2z+1

2x+z=0

z=-2x 

x+5y+z=5

x=5-5y-z 

z=-2(5-5y-z)

z=-10+10y+2z 

z+10=10y+2z

10=10y+z

z=10-10y

x=5-5y-(10-10y)

x=5-5y-10+10y

x=5y-5

y=5y-5+2(10-10y)+1

y=5y-5+20-20y+1

x=0

y=1

z=0

y=-15y+16

16y=16

y=1

z=10-10y

z=10-10

z=0

x=5y-5

x=5-5

x=0

q = -10.5 

-4q = 42

/4   = /4

  q= -10.5

-4q=42

q=42/-4

q=-10.5

x = 0 stop asking stupid questions dude thats like 1 + 1..... It's 2 if you didn't know.... -_-

Dude 5x5 = 25 you should know that!

Don't ask stupid questions...

This isn't a math problem!

We got a calculator just type it in that 

This is without tan so you know!

((70cos45)/(130+70sin45)) = 0.2757558275985764

20/3 = 6,66666666666667 

6,66666666666667 ≈ 6,7

area of circle

x=0

over 9000!!11!1!!

\(\lim \limits_{x\to 0} { \left( \frac{ \sin{(x)} } {x} \right)^{ \cot{(x)} } }= \ ?\\ \)

Part 1:

We need

\(\boxed{~ \begin{array}{rcl} \cot{(x)} &=& \frac{1}{x} -\frac13 x -\frac{1}{45} x^3 -\frac{2}{945} x^5 -\frac{1}{4725} x^7 -\cdots \\ \ln{(1+x)} &=& x -\frac{x^2}{2} +\frac{x^3}{3} -\frac{x^4}{4} +-\cdots \\ \hline \sin{(x)} &=& x -\frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} +\frac{x^9}{9!} -+\cdots \\ \sin{(x)}-x &=& -\frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} +\frac{x^9}{9!} -+\cdots \\ \frac{\sin{(x)}-x}{x} &=& -\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \\ \hline \frac{\sin{(x)}}{x} &=& 1+ \left( \frac{\sin{(x)}}{x} - 1 \right) = 1+ \frac{\sin{(x)}-x}{x} \end{array} ~}\)

\(\boxed{~ \begin{array}{rcl} \ln{(1+z)} &=& z -\frac{z^2}{2} +\frac{z^3}{3} -\frac{z^4}{4} +-\cdots \\ z &=& \frac{\sin{(x)}-x}{x} = -\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \\ \ln{ ( 1+ \frac{ \sin{(x)}-x } {x} ) } &=& \left( -\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^1 \\ &&-\frac12 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^2\\ &&+\frac13 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^3\\ &&-\frac14 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^4 \end{array} ~}\)

continued...

I'm NOT familiar with the description of triangle in terms of letters. I assume the the given side "TO" is the Adjacent side.If that is so, then we have:

Cos(60)=5/Hypotenuse

.5=5/H

H=5/.5

H=10, then we have:

10^2=5^2 + (HT)^2

HT^2=100 - 25=75

HT=Sqrt(75)

HT=8.66. If "TO" is the Hypotenuse, then:

HT=4.33

4 * (6*2) = 4*12= 48

Use the key "x^y". Example: 2^3. You enter 2 first, hit the x^y key followed by 3 and then = sign.

Sure. As many nuts as in your head!!!!!

Liliana takes 4 breaths per 10 seconds during yoga. At this rate, about how many breaths would liliana take in 2 minutes of yoga. Explain.

4 / 10=.4 breaths per second that Liliana takes during yoga, therefore in 2 mintutes,

(2 X 60) seconds X .4=48 breaths that Liliana will take in 2 minutes.

He's copying from an online AoPS class.

15%  ......there is an implied decimal point after the "5"

Drop the "%"  sign and move the decimal two places to the left  = .15

Another example  :   23.6%   =   .236

Finally :   .006%   =   .00006

Hey.....don't get too cocky........

I know where you live   !!!!!

9+10=19

Note that we are taking the limit as x approaches 0 from the right.......

Using the Puiseux  series expansion for the first six terms of sin (sqrt(x)), we have :

lim x →  0   [sqrt(x)-x^(3/2)/6+x^(5/2)/120-x^(7/2)/5040+x^(9/2)/362880-x^(11/2)/39916800] / x

Dividing each term by x   and taking the limit as x → 0  on the last five terms will result in 0

Dividing the first term by x , we have

lim x →  0      1 / sqrt(x)

As x → 0,    the denominator gets extremely "small" in comparison to the numerator .... so the function  →  +infinity  as x  → 0 ....i.e., .......the limit does not exist

Here's the graph that confirms this  .....   https://www.desmos.com/calculator/tzfgdcaps7

In a triangle, the base is 4 cm. If the area of the triangle is 30 square cm, what is the height (in cm) of the triangle?

Area of Triangle=(Base X Height) / 2

30=(4 X H) / 2

30=4H/2 cross multiply

4H=60

H=60/4

H=15cm. The height of the triangle.