All Answers 358514 Answers

All the x, y values on or "above" the line

Note that the equation shown, -1.2x + 2y = 1.6, can be transformed into yours

Multiply everything by 2.5

-3x + 5y = 4     Rearrange

5y = 3x + 4

But, since all the points we're interested in lie on or above the line, we have

5y ≥ 3x + 4   

There you go anonymous.

You got lots of attention and maybe some people learned about the OEIS foundation along the way!

$$f^{-1}(x)=\{(4,3),(3,4),(6,-2)\}$$

I think that is correct 

yes there is although some ofthe features do not work properly on my android.

The system is being overhauled.  There are some bugs that have to be worked through!

Another great graph Chris, thank you.

Yes fair enough - there was not a question - sorry.

deleted deleted 

Thank you Chris

Thanks Chris,

What an impressive sketch! 

Since there is no equal sign, the problem cannot be solved. Also, it is impossible to simplify the equation because in "m * + m" there is an operator immediately followed by another operator, which is not possible in mathematics.

$$0.04=\frac{4}{100}=\frac{1}{25}$$

$$\sqrt{68}=\sqrt{4\times 17}=\sqrt4\times\sqrt{17}=2\sqrt{17}$$

or you can get an approximation off a calculator.

Since we are subtracting a positive number from a negative number in this equation, we need to apply a rule called same-change-change. This is where the first number of a problem stays the same, but the operator and the second number change. (For example, "+" can become "-" and "-8" can become "8".) This turns:

-51 - 6

into:

-51 + (-6)

Now we are adding -51 and -6, which gives us -57. Thus, -51 - 6 = -57.

OR

you could say that there are 52 weeks in a year 

so in a year you would earn 400*52= $20800

then there are 12 months in a year 20800/12= $1733.33 per month

For some reason your graph is not displaying Chris.

Sometimes pictures wander around the place like in hogwart's school.  

It might wander back later. 

Not quite infinty but a ridiculously big number. 

What is the ? again????????

Rather than checking all the above answer (which was done by substitution) I am going to d it a different way (by elimination) and see if the answers are the same.  If they are then i will check off both.

x-2y=24 (1)

x-y=10 (2)

(1)-(2)

-2y--y=24-10

-y=14

y=-14

----------

Sub back into (2)

x--14=10

x+14=10

x= -4

Good they are the same!  

I would have skipped the first stem and gone straight to 0.6*144cm

But it makes no difference.

Which means you do $$7^5$$ first (because the 5 is an exponent) and then you multiply by the number afterwards.

The deceleration force, F,  is F = 0.183*mass*g (g is acceleration of gravity).  This equals -mass*a, where 'a' is acceleration (deceleration) of the car, so

a = -0.183*9.81m/s². (The negative sign just means it is slowing down.)

With constant acceleration we have the kinetic equation v² = u² +2as, where v = final velocity (0), u = initial velocity (51.8km/hr = 51.8*10³/3600 m/s) and s is distance travelled.  So:

0 = (51.8*10³/3600)² - 2*0.183*9.81*s

 s = (51.8*10³/3600)²/2*0.183*9.81

$${\mathtt{s}} = {\frac{{\left({\frac{{\mathtt{51.8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{3}}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.183}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{57.663\: \!954\: \!885\: \!109\: \!463\: \!3}}$$ metres

or s ≈ 57.7 metres

The force needed to get the cup to slip is 0.42*cupmass*gravitational acceleration. This must equal cupmass*accelerationof plane. Therefore:

acceleration of plane = 0.42*9.81 m/s² (assuming plane isn't so high gravitational acceleration is significantly reduced!), as the mass of the cup cancels out.

 $${\mathtt{AccnOfPlane}} = {\mathtt{0.42}}{\mathtt{\,\times\,}}{\mathtt{9.81}} \Rightarrow {\mathtt{AccnOfPlane}} = {\mathtt{4.120\: \!2}}$$ m/s²

The resistive force is given by μ_s*W, where W is the weight.  The resistance is overcome a force of 659.2N is applied, so 

μ_s*87.9*9.81 = 659.2

μ_s = 659.2/(87.9*9.81)

$${\frac{{\mathtt{659.2}}}{\left({\mathtt{87.9}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} = {\mathtt{0.764\: \!468\: \!009\: \!356\: \!383\: \!3}}$$

or μ_s ≈ 0.764