All Answers 354302 Answers

Only at c = 0   is there 6 solutions

the only place an integer line of  'y' crosses the graph in six places is at y = 0

Notice that 2ⁿ for n = 1 • • •  the value of 2ⁿ is as follows: 

      2,         4,         8,       16,

    32,       64,     128,     256,

  512,   1024,   2048,   4096,

8192, 16384, 32768, 65536,

etc. 

The units digit repeats after every four exponents. 

Example:  Say you wanted to know 2⁷ ... divide the 7 by 4 and the remainder is 3. 

                So the units digit will be the third one in the repeating groups, which is 8. 

Example:  Or, you wanted to know 2¹⁶ ... divide the 16 by 4 and the remainder is 0. 

                But when a remainder is 0, that's the same as it being the same as the divisor, which was 4. 

                So the units digit will be the fourth one in the repeating groups, which is 6. 

Okay, skip ahead to 2²⁰¹⁰ ... divide the 2010 by 4 and the remainder is 2. 

That means the units digit will be the second one in the repeating groups, which is 4. 

_.

To be divisible by three the sum of all of the digits has to be divisible by 3 also

3+4+1+4+x+7 = 19+x        if x = 2 5 8    it will be divisible by 3  

   you can finish the Q.........

Dividing ANYTHING by zero is UNDEFINED....NOT ALLOWED

Compute the differences between the serial terms

3y          7           15            2y+1

      3y-7        -8          -2y+14                  then compute the differences between these serial terms ....they should all be equal

             3y+1        2y -22                        these two terms are equal

3y+1  = 2y-22

y = -23                                (If I did this correctly....I just did a quick lesson on Quad Sequences) 

Commision is 11%

.11 x 5300 = $ .......

   multiply this answer by 52 to get a yearly total (there are 52 weeks in a year)

yeah i want to know how to do it thx though 

w a i t...

i think im wrong 

For one, we know that x cannot equal -2 because f(-2) is either undefined or indeterminate. 

To find the numbers that aren't in the domain of g(x), we set the bottom to 0. 

So 3x^2 + 11x + 10 = 0

(3x + 5)(x + 2) = 0,

x = -5/3, -2

-2 + (-2) + (-5/3) = -4 - (5/3) = -17/3

Uhm I'm not very good at functions and this stuff so there's a lot of room for errors :/. Please correct me if I'm wrong.

Really?!?  What's your answer?

The answer to this question is:      sin∠PAQ = 0.6   or   3/5

btw, your above post is ridiculous

Why do you need all that to solve such a simple question?!!

x + 15 = 3x + 5

2x = 10

x = 5

NM = OL = 5 + 15 = 20

The answer is "B" or the second one.

Please correct me if I'm wrong. 

The probability is 29/48.

By calculus,

\([\mathcal{R}_1] = 30 + \frac{1}{4} \cdot \pi \cdot 10^2 + \int_0^5 \sqrt{100 - x^2} \ dx + \int_0^6 \sqrt{100 - x^2} \ dx.\)

We can write out the areas similarly, to get [R_1] - [R_2] - [R_3] + [R_4] = 80.

Total distance is 5 + 5 km = 10 km

t_u =  5 / ( 30-12)  = 5/18

t_d = 5 / (30+12)   = 5/42

total distance / total time = average km/hr  

    10 /( 5/18 + 5/42)  = 10 / (300/756)  = 7560/300 = 25.2 km/hr

Let the time it took her to walk home =T

8T =48[T - 20], solve for T
T = 24 minutes it took her to walk home.
24/60 x 8 km/h =3.2 km - distance between car shop and her home(or how far she walked).
24 - 20 = 4 minutes it took her to drive to car shop.
4 minutes + 10 minutes she stayed there + 24 minutes to walk home =38 minutes after noon hour when she arrived home.

Alan,

thanx, I've got this one., very kind of you..

Alan,

I really do not know....I'm thinking something like this:

\({x \over y}=-1\)

so then \(x=-y\)

so then \(x+y=0\)

However, it is given that \(x+y=2\)

uuhh, Alan...I don't know...this makes me feel so utterly stupid!

Hi Alan,

gosh, I don't even have a slightest clue where to start...how to start....I'm looking at this..but just do not know at all what to do..I have just googled..found something..maybe.....just give me a minute or so, I'll come back to you..

thank youuuuuu!!!!!!!!!!

Part 1 is as follows:

Let's see you attempt part 2.

As follows:

You CANNOT make 69 confusions out of 11 and 8 confusions !! In fact, it is the largest number that you CANNOT make using 8 and 11 confusions.

2-1=3

45