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If the x-axis provides a line of symmetry, then there will be a point below the x-axis for every point above the x-axis and as far below as the other point is above (and vice versa).

If you start at (5, 2) and drop straight down through the x-axis as far as this point is above the x-axis, you will end at (5, -2).

These are the numbers 3, 7, 13, 17, 23, 27, ..., 93, 97, 103, 107, ...

I'm going to divide these into two groups. First, the ones that end in 3.

Now, I'm just going to add the first few:

          3 + 13 + 23 + 33 +  ... + 93.

If you use the formula for the sum of an arithmetic series:  Sum = N(F + L)/2

  N  =  number of terms = 10          F = first term = 3          L = last term = 93

--->   Sum  =  10(3 + 93) / 2  =  480

Now, I'm going to add these:  103, 113, 123, ..., 193

--->   Sum  =  10(103 + 193)/2  =  1480

     (Notice that this sum is 1000 greater than the previous sum because each of the ten numbers is 100                greater than the previous ones and 10 x 100 = 1000.)

Now, to add:  203, 213, 223, ..., 293:  their sum is 2480.  (Use either the formula or just add 1000 to the previous answer.

Adding 303, 313, 323, ..., 393:  their sum is 3480

Etc.

Adding 903, 913, 923, ..., 993: their sum is 9480.

We end with these sums:  480 + 1480 + 2480 + ... + 9480.

Add these together -- by calculator, or again use the formula.

This gets you halfway home -- you still need to find the sum of 7, 17, 27, 37, ..., 997.

You can do this in the same manner that was used to find the sum of 3, 13, 23, 33, ..., 993

   or you can realize that each of these numbers is 4 larger than the numbers of the first set   ---   if you can      figure out how many number end with 7, you can take that number, multiply it by 4 and add that to the          answer for the numbers ending with 3.

Finally, add the answer for the numbers ending with a three with the answer for the numbers ending with a seven.

Try using the square rooot function on most cauculators.

the whole thing

-7d + 8 < 29

Subtract 8 from both sides:

-7d < 21

Divide both sides by -7  ---   However, when you multiply or divide both sides of an inequality by a negative number, you must change the direction of the inequality. You do not change the direction if you multiply or divide by a positive, or if you are adding or subtracting:

  d > -3

What is the question?

Which parts?

A formula for the n^th term of a geometric sequence is:  t_n  =  a · r^{n - 1}.

--->   16  =  a · r²² 

--->   24  =  a · r²⁷

Dividing the top equation by the bottom equation (and cancelling out the a's):  16/24  =  r²² / r²⁷

--->  2/3  =  r^-5     --->   3/2  =  r⁵     --->   r  =  (3/2)^1/5     (which is the fifth root of 3/2)

Since  16  =  a · r²²      --->   16  =  a · ( (3/2)^1/5 )²²    --->   a  =  16 / ( (3/2)^1/5 )²²  

To find the 43rd term:  a · r^{n - 1}     --->    16 / ( (3/2)^1/5 )²² · ( (3/2)^1/5 )⁴²    --->   16 · (3/2)⁴

please help me, I dont understand

The value of x, which was the number to be added to each of the three original numbers, is 100.

wait, what would the answer be?

i am not sure, but it says that this answer is wrong.

Your instructions say to add the same number to 60, 100, and 150.

If I add 100 to each of these I get:  160, 200, and 250

The ratio of 160 to 200 is 160/200 = 0.80

The ratio of 200 to 250 is 200/250 = 0.80

They have the same ratio, thus, they form a geometric sequence.

What is the answer supposed to be?

The formula for the n^th term of an arithmetic sequence is:  t_n  =  a + (n - 1)d

--->  2/3  =  a + 22d

--->  3/2  =  a + 52d

Subtracting the top equation from the bottom equation: 5/6  =  30d   --->   d =  1/36

Multiplying the top equation by 52:       104/3  =  52a + 1144d

Multiplying the bottom equation by -22:    -33 = -22a - 1144d

Adding down:  5/3  =  30a   --->   a  =  1/18

To find the 35th term:  value  =  1/18 + (35 - 1)(1/36)     <---  the answer.

Let the constant be x:

By the definition of a geometric sequence, then:  (60 + x)/(100 + x)  =  (100 + x)/(150+ x)

Cross multiplying:  (60 + x)(150 + x)  =  (100 + x)(100 + x)

--->   9000 + 210x + x²  =  10000 + 200x + x²

--->   10x  =   1000

--->   x = 100 

Thanks Melody,

At the moment I'm just doing any online tests at the GCSE (senior/high school)/University entry level and anything I have trouble with, I dig into until I have a better understanding of it or until my brain starts to fry, then I go back to the tests to find my next challenge. I'm at the brain frying point with factorising at the moment so I'm letting what I have learned so far sink in but this one turned up in a sample test and I felt it was at a level that I should have been capable of so couldn't resist challenging it.

I'm trying to cover everything in maths up to a similar level, going up in increments across the board so I don't get to confident in some areas at the expense of others. I don't see this being a quick task but it should be worth it.

Thanks.

The formula you require for this is time  = distance/speed.

The formula for distance , speed and time calculations is $${\frac{{\mathtt{D}}}{\left({\mathtt{S}}{\mathtt{\,\times\,}}{\mathtt{T}}\right)}}$$ where D=Distance, S=Speed and T=Time. Whichever answer you require, cover it and the remaining is the formula for that answer. So if i wanted to know the distance, I would cover the "D", leaving the formula "S" multiplied by "T". If I wanted to know the time as in your question, I would cover the "T", leaving "D" divided by "S".

I find it helps me to remember this by the fact that the letters appear in the order they appear in the word "DiSTance", the three letters don't all appear in either "Speed" or "Time".

It is important to ensure that you are using the same units of measure for the distance and speed e.g. Miles for distance and MPH (miles per hour) for speed, or Kilometres for distance and km/h (Kilometres per hour) for speed. It doesn't matter what unit of measure you use, as long as you use it across the whole calculation. If you use any non metric unit of measure you must convert any fraction of a whole unit into a decimal, though this has already been done on your question.

As the question hints towards, you also need to ensure that the time applied to the calculation is expressed in decimal hours e.g. 15 minutes = 0.25 hours, 1 hour 45 minutes = 1.75 hours .etc. As your question gives the speed to a whole hour with no fraction, you don't need this for this question. You may then need to convert this back to the conventional hours and minutes if required.

There is no need for me to go into the calculations of the particular details of you question as geno3141 has already given you this and is a more reliable source than myself, I'm just a newbie. I have just laid this out as simple as possible in case it is required this way and to help with further Distance, Speed, Time calculations.

The epression is 1337(-½)ⁿ and you want this > 1

The negative sign flips the answer from negative to positive, to negative, etc.

I'm going to ignore the negative sign, but realize that my answer will be two times too large, because it will include both the positive and the negative answers.

1337(½)ⁿ  >  1   --->   (½)ⁿ  >  1/1337   --->   2ⁿ  <  1337   --->   n = 10

But, this contains both the negative and the positive answers, so, removing the negative solutions,  n = 5.

What two answer? Sorry. Thank you so much 

You can use the sum of an arithmetic series to find the sum of: 3 + 13 + 23 + ... + 93  =  480

Now, to find the sum of the aritmetic series: 103 + 113 + 123 + ... + 193:  there are 10 of these numbers and each number is 100 greater than the number in the previous series, so their sum is 10 x 100 + 480  =  1480.

Similarly, 203 + 213 + 223 + ... + 293  =  1000 + previous answer  =  2480.

...

903 + 913 + 923 + ... + 993  = 9480.

You can use the sum of an arithmetic series to find the sum of these answers:  480 + 1480 + 2480 + ... + 9480.

Now, for those numbers ending in 7:  each of them is 4 more than the corresponding number ending in 3. So, to the answer for those numbers ending in 3, we must add 4 x the number of numbers ending in 7.

Add these two answers together.