All Answers 358090 Answers

I just read this post and I thought it was really, really funny!

Those pics are awesome melody!!! Lol, Zegroes:)

Yes, three people are online.

Do u think it just might be possible to plug it into a calculator?

anyone  online?

How long does it take for Marks car to change to its velocity from 12 m/s to 25 m/s if the acceleration is 5 m/s2

a = acceleration

v = velocity

t = time

\(\begin{array}{rcll} \boxed{~ \begin{array}{rcll} a &=& \frac{ \Delta v } { \Delta t } \end{array} ~}\\\\ \Delta v &=& 25\ \frac{m}{s} - 12\ \frac{m}{s} \\ a &=& 5\ \frac{m}{s^2} \\\\ \end{array}\\\\ \begin{array}{rcll} 5\ \frac{m}{s^2} &=& \frac{ 25\ \frac{m}{s} - 12\ \frac{m}{s} } { \Delta t } \\ 5\ \frac{m}{s^2} &=& \frac{ 13\ \frac{m}{s} } { \Delta t } \\ \Delta t &=& \frac{ 13\ \frac{m}{s} } { 5\ \frac{m}{s^2} } \\ \Delta t &=& \frac{ 13 } { 5 } \ s \\ \mathbf{ \Delta t } &\mathbf{=}& \mathbf{2.6 \ s }\\ \end{array}\)

How did you find the f(x) to substitute in the first place?

It will equal the change in kinetic energy:   (mass/2)*(v² - u²)   where v = final velocity (4m/s) and u = initial velocity (3m/s).

With mass in kg the result will be in Joules.

.

v = u + a*t   where v = final velocity (25m/s), u = initial velocity (12m/s), a = acceleration (5m/s²) and t = time (secs)

So t = (v - u)/a  

I'll leave you to fill in the numbers.

R u still on jc?

Noooo! I missed u jc!?!

Just calculate gms sug/ounce cereal for each

Frosted Flakes  11gms/1oz  = 11 gm per oz

Raisin Bran        13     / 1.4   = 9.29 gm per oz     Frosted Flakes has more gm sugar per ounce (as expected!)

Guess I should say RAISIN BRAN has less sugar (since THAT is what the Q asked to determine)

   and I had  gm and oz reversed originally....

Emma invested money in a local stock. The money can be modeled by using the expression P=10,000(1.1)^t where t is the time in years and P is the amount of money, in dollars. How much money did she have three years ago? Round your answer to the nearest dollar

Emma had a total of $7,513.15 three years earlier.

Uhhhh...I'll need a diagram or more information to answer anything about this .... OK?

A sector of a circle is inside a rectangle. The rectangle has the side lengths 25cm and 15cm. This also means that the radius of the sector is 25cm. What would be the area in the rectangle that is outside the sector?

The answer is 185cm^2 but i dont no how to get it. Please help!

 \(\begin{array}{rcll} A_{\text{rectangle}} &=& 25\cdot 15 \ cm^2 \\ &=& 375 \ cm^2 \\\\ A_{\text{sector of a circle}} &=& \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2 \\ A_{\text{Area in the rectangle and outside sector}} &=& A_{\text{rectangle}} - A_{\text{sector of a circle}}\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{\varphi}{360^\circ} \ cm^2\\\\ \sin{(\frac{\varphi}{2})} &=& \frac{\frac{15}{2}}{25} \\ \sin{(\frac{\varphi}{2})} &=& \frac{7.5}{25} \\ \sin{(\frac{\varphi}{2})} &=& 0.3 \\ \frac{\varphi}{2} &=& \arcsin{(0.3)} \\ \frac{\varphi}{2} &=& 17.4576031237^\circ \\ \varphi &=& 2\cdot 17.4576031237^\circ \\ \varphi &=& 34.9152062474^\circ \\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{34.9152062474^\circ}{360^\circ} \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - 190.432908760 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 184.567091240 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &\approx& 185 \ cm^2\\\\ \end{array}\)

7n/20 as a percent in terms of n

7n/20 = n  7/20

as a percent    n  7/20 x 100 = 35(n) %

~jc

You can MULTIPLY fractions together without finding a common denominator.    To ADD or SUBTRACT you WILL have to find the common denominator

 A Common denominator for 2 and 3  is...........    '6'

So, convert them both to 'sixths'

1/2  x 3/3  =  3/6

1/3  x 2/2  = 2/6

Substitute   1/2 - 1/3 =  3/6 - 2/6  =    1/6

~jc

In this problem t = -3    (for time THREE years ago)

Just substitute -3 in for t  to get your answer.....

= $7513

~jc

Yikes !   That is probably more than you wnated to know (or I remember from advanced applied math years ago)   I'll believe the algorithm Guest#2 wrote about is what the hardware executes to determine the value for sin.

     The important thing to remember is the Sides of the RIght triangle determine the angle between them....that angle will always have the same value for sine cos etc.

......and the 'calculator'  just 'calculates' the value for this angle.....

Thanx Guest #2......wish I remembered that stuff !

Well .... that equals   (3x-6)(3x-6)    

If you multiply it out     9x^2 - 18x - 18x + 36 = 9x^2-36x +36     or   9 (x^2- 4x +4 )

~jc 

1/4 of it !     Ha ha

1/4 of 1  3/5    is   1/4 x 1 3/5 =  1/4 x 8/5 = 8/20 = 2/5 ths lb left over. (IF he ate what he cooked!)

~jc

e = one diagonal

f = the other diagonal

h ist the height of the rhombus

\(\boxed{~ \begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }} \\ \end{array} ~}\\\)

or

\(\begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }}\\ h &=&\frac{ 1 }{ \frac { \sqrt{ e^2 + f^2 }}{ e\cdot f}} \\ h &=&\frac{ 1 }{ \sqrt{ \frac{e^2}{e^2\cdot f^2} + \frac{f^2}{e^2\cdot f^2} }} \\ h &=&\frac{ 1 }{ \sqrt{ \frac{1}{f^2} + \frac{1}{e^2} }} \\ \frac{1}{h} &=&\sqrt{ \frac{1}{f^2} + \frac{1}{e^2} } \\ \frac{1}{h^2} &=&\frac{1}{f^2} + \frac{1}{e^2} \\ \frac{1}{h^2} &=&\frac{1}{e^2} + \frac{1}{f^2} \\ \end{array}\)

\(\boxed{~ \begin{array}{rcll} \frac{1}{h^2} &=&\frac{1}{e^2} + \frac{1}{f^2} \end{array} ~}\)

4 minutes is  4 x 60 seconds = 240 seconds

Nadia stops 12 seconds before the music ends  so   240 - 12 = 228 seconds she dances

Percentage of time she dances is   228/240 x 100 = 95%

~jc