All Answers 358782 Answers

most likely 2

the 2 is significant and so is the 3.

BUT it really does depend on the situation.  You need the whole question to say for sure.  :)

$$0.861m^2 + 0.4m^3$$

This cannot be simplified.  $$m^2 \;\;and\;\;m^3$$      are not like terms.  Like terms have to have the same letters and they must be raised to the same powers.

You made a caresless error anon.  So it is not quite right :(

$$\\\frac{30}{140}\times\frav{1}{6}\\\\
=\frac{3}{14}\times\frac{1}{6}\\\\
=\frac{1}{14}\times\frac{1}{2}\\\\
=\frac{1}{28}$$

Let x= cos (theta)

Can you do it now?

-11 divided by 2/4 = -11 x 2

∴ -11 divided by 2/4 = -22

(30/140) x (1/6)

30/140 becomes 3/14

(3/10) x (1/6)

to multiply fractions, you multiply the numerators together and the denominators together.

(3x1)/(10x6)

3/60 <=> 1/20

the answer is 1/20

The second elevation is irrelevent

The third elevation is 94 feet higher than where he began = -418 + 94 = -324  feet

If he descends another 132  feet, we have   -324 - 132 =  -456 feet

Looks like you're OK with your answer  !!!!

the answer is y: (5,-15)

the x goes up by 2, so we do another jump of 2 to reach 5

the x drops by 10 so we do another drop by 10 to get to -15

We have

[1 + x]/2 = 3                 and                               [ 5 + y] / 2  = -5

                   Multiply both sides of each equation by 2   

1 + x  = 6                                                           5 + y = -10

subtract 1 from each side                                   subtract 5 from both sides

x = 5                                                               y = -15

So....the other point is  (5, -15)

If f(x) = x^2 + 5    and g(x)  = √(1 - x)....then you already have (f/g)(x)

I'll give this one a shot.......

Let the larger number be :

a10^4 + b10^3 + c10^2 + d10  + e

And let the smaller number be :

e10^4 + d10^3 + c10^2 + b10 + a     

And subtracting the second from the first, we have

(a - e)10^4 + (b - d)10^3 + (d - b)10 + (e - a)     rewrite as

(a - e) 10^4 + (b -d)10^3 - (b - d)10 - ( a - e) =

(a - e)(10^4 - 1) + (b - d)(10^3 - 10) =

(a - e)(9999) + 990(b - d) =

(a - e)(3^2 x 11 x 101) + (b - d) (2 x 3^2 x 5 x 11) ]

11 [ (a - e)(3^2 x 101) + (b - d)(2 x 3^2 x 5) ]

And since a,b,c,d,e are random digts between 0 and 9 inclusive, neither (a - e)  nor (b - d) can be any prime number greater than 7

So it appears that 11 is the largest prime number that must be a factor of this number

-------------------------------------------------------------------------------------------------------------------

Can some other mathematician see if this is correct???

I think you need a f and a g function first :/

Net profit = gross profit - costs =  money in - costs

There are no p's in the original question.   

There's nothing to "solve"......we need an equal sign somewhere for that......

172 (60m) = 8

10,320m =8

m=8/10,320

m=1/1290

(1, 0.6) and (2, 1.2)

m = slope = [ 1.2 - 0.6] / [ 2 - 1 ] =  [0.6]/ [1]  =  0.6

Find slope between (1, 0.6) and (2, 1.2)

x-intercept | 0.
y-intercept | 0.
slope | 0.6

Thanks, Melody..it was a "bear" to calculate all of that !!!...I'm not totally sure about my answer, but it seemed to "work".....maybe someone else will have a different "take"......????

This looks like it was a doosey of a question Chris ;)

Thanks, Melody......I'm not sure that this answer is unique....for instance, if triangle ABC had a different shape [but the same area as above], would triangle AEF still have the same area???......I don't really know.....maybe someone else could prove/disprove it.....!!!!

powers are done before3 multiply so

$$99*\frac{1}{3^x}=\frac{99}{3^x}$$

Nice pic Chris ;)

x-intercept | 0.
y-intercept | 0.
slope | 0.4

NO it would rarely be true,

it would depend on what a and b are. (Assuming that they are constants)