Questions on Algebra: Decimal numbers, power of 10, rounding answered by real tutors!

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All of your problems are sums of arithmetic sequences.
My problem is that I do not know what strategy you were expected to use to solve them. My way of solving them may not be what your teacher expects. You may be expected to look up a formula in your textbook and apply it, and the calculation may be different from the one I would do (same result, of course).
Your textbook or class notes may tell you that an arithmetic sequence can be represented as
, , , .... , ...
with
with the values of and determining the sequence.
Then they could have said that
.
The way your problems read, you are given , , and , so that formula will work well.
Another formula you could have been given is

That one is the easiest to derive for me (I don't memorize formulas; I just derive them from the definitions as needed). It would be easiest to use if you had the first and last terms already calculated.
You may have been given some other equivalent expression.
My best guess is that you are expected to use the first formula above, so I'll go with that.
51.
(You could also say , and , so
, using the second formula).
52.
53.
54.
55.
56.
Because I do not memorize formulas, I would have calculated it like this:
, splitting the sum into other sums, to make it easier to use the formula that makes more sense to me, but I bet that's not what your teacher expects.

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The trick is to use the distributive principle.  Then use the 
special formulas:

1. You can distribute ∑ over a sum or difference of terms.
2. You can take out a constant in front of the ∑.

I'll just do this one and you can easily do the rest.

51. 20 on top of symbol, i=1 on bottom,(3+5i) on right.

First distribute the ∑ over the sum like this:

 + 

Write the 3 as 3·1

 + 

Now you can take the constant 3 out in front of the first 
summation and take the constant 5 out in front of the second
summation, like this:

 + 

Now you can use the special formulas:

 +  =  +  = 60 + 1050 = 1110.

Now you can do all of them by using the distributive principle first
and then using the special formulas.

Edwin

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Since the product of two of these integers is a power of 2, it follows that both integers are powers of 2. Hence these integers can be {1, 2^2011}, {2, 2^2010}, ..., {2^1005, 2^1006} (order doesn't matter). The two integers cannot be {1, 2^2011} since the third integer would be zero. However, each of the other ordered pairs produces a unique integer (e.g. subtract the sum of the two integers from 2^2011 + 1), so the number of values for the third integer is the number of ordered pairs in {2, 2^2010}, ..., {2^1005, 2^1006}, or 1004 integers.

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0.4
4/10
=2/5

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Okay I think what you are asking is to simplify




We have an exponent rule that says something like this:


So all we are doing is using this and a few other tricks

can be broken up as follows

Now 3/15=1/5
and
is just
and using the rule above this is just
So from

we get



We could also use the rule

to write our answer as


Hopefully this helps, please let me know if you are unsure:)

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( rounded to nearest tenth )
This problem gets into the subject of
" significant figures "
Actually, not all the digits after the decimal point are
really " significant ".
Notice the hours is 86 and not 86.3, or 85.9, so
86 is as close as can be judged
That is 2 significant figures.
3543 has 4 significant figures. They could have said
3543.27. That would have been 6 significant figures
--------------------
The number of significant figures in the answer is
determined by the least accurate number in the input data,
so really only 2 figures in answer can be relied on.
So 3 figures is just fine

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If Lynn can type a page in p minutes, what piece of the page can she do in 5 minutes?
----
Use a proportion:
piece/time = piece/time
-----------
x/5 = 1/p
------
x = 5/p of a page
========================
Cheers,
Stan H.
========================
A. 5/p
B. p - 5
C. p + 5
D. p/5
E. 1- p + 5

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numbers are given
take amount of entertainment over total amount of take home

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no numbers in the problem. known: and and im guessing from the statement so











this was a bit tricky at first i didnt think there was enough information but i stuck with it and amazed myself.

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a positive exponent moves decimal point to the right as many as the exponent indicates
a negative exponent moves decimal point to the left as many as the exponent indicates
=

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fyi this isnt a decimal problem
ex:
same concept below

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2*2*2*2=16
16-8=8
8/4=2

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trial and error wins agian
start with 99^2
and then 98^2
97^2
...
88^2=7744
so a=7 b=4
could be more answers but u asked a yes or no question and the answer is yes

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theres no equation...

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ex: .100>.099
best to list them all below to visually see them
.067
.070
.600
.760
now i added the 0's at the ends to show places to visually guide you to which is greater than others
if you took the decimal point away it would look like
67
70
600
760
so now you should easily see which is the greatest to the least

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its already rounded to the 10th what are u asking?

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.5 to fraction

--->reduced
.53 to fraction


.546 to fraction

--->reduced

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.09 will never = 1
.09+.91=1
.09/.09=1
.09*10+.1=1
.09*100-8=1
but .09 will never = 1 by itself

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Factor 55^4 into the product of prime factors by first factoring 55 and then raising each of its factors to the fourth power.
:
(5*11)^4 = 5^4 * 11^4 = 625 * 14641 = 9,150,625

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1. Convert the binary number 111000111 into the base ten equivalent.

 1     1      1      0     0      0      1     1      1 

1*28 + 1*27 + 1*26 + 0*25 + 0*24 + 0*23 + 1*22 + 1*21 + 1*20
 258 +  128 +  64  + 0  +  0  +   0  +   4  +  2   +  1   = 457

2. Convert the base ten number 111 into the binary equivalent.

Keep dividing by 2 putting the quotients underneath and the
remainder out to the right beside the quotient.  Quit when
the quotient becoms 0.  Then the digits of the binary number
are the remainders reading from bottom to top:

   2)111
    2)55  1
    2)27  1
    2)13  1
     2)6  0
     2)3  1
     2)1  1
       0  1    
   
List the remainders from bottom to top

Answer:  1110111 


3. Convert the decimal number 0.35743574357435743574357435743574...
   to a fraction

       N = 0.35743574357435743574357435743574...

There are four repeating digits so multiply both sides by 104 or 10000

 10000N = 3574.3574357435743574357435743574...

Subtract the original equation for N

 10000N = 3574.35743574357435743574357435743574...
      N =    0.35743574357435743574357435743574...
--------------------------------------------------
  9999N = 3574

      N =   It doesn't reduce.   


Edwin

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suppose you take the number
2187
factorize the number by prime factors (2,3,5,7,11,13,17,etc
keep on dividing . 2187/3 = 729. 729/3=243
3 | 2187
3| 729
3{243
3|81
3|9
3|3
3|1
you have 3*3*3*3*3*3 = 3^6
Same can be applied to any number
m.ananth@hotmail.ca

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There are no 2's in binary.
------------------
If the 70 & 80 are Hex numbers:
70 = 0111 0000 --> 1's comp = 1000 1111 --> 2's comp = 1001 0000
80 = 1000 0000 --> 1's comp = 0111 1111 --> 2's comp = 1000 0000
-------------
If they're decimal:
70 = 46H = 0100 0110 --> 1's comp = 1011 1001 --> 2's comp = 1011 1010
80 = 50H = 0101 0000 --> 1's comp = 1010 1111 --> 2's comp = 1011 0000

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120.00 - 50.75

You can subtract in two steps:
1.00 - 0.75
and 119 - 50

1.00 - 0.75 = 0.25

and 119 - 50 = 69



And then sum: 69 + 0.25

The answer is 69.25


Any question, please ask me. My email is AREVALOGABRIEL2@GMAIL.COM

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To convert a repeating decimal to a fraction

Step 1: Count the number of repeating digits. This is n.

Step 2: Set x = the decimal number you're trying to convert.

Step 3: Multiply both sides by 10^n where n - the number of repeating digits.

Step 4. Subtract the original equation from the new equation.

Step 5. Solve for x.

Step 1. there is one repeating digit (9) so n=1.

Step 2. x=99.999...

Step 3. 10^1 = 10 so 10x=999.999... (we can add as many 9's as needed on the end because it repeats.)

Step 4. 9x = 900

Step 5. x = 900/9 = 100.

It seems wrong, but the fact is .999999999... = 1. So 99.999999999... = 100 and that's 100/1 as a fraction.

42.242424 has 2 repeating digits (24). So n=3.

x=42.2424242

10^2 = 100 so it's 100x = 4224.2424242 (add as many repeating digits as necessary so when subtracting, the decimal portion is gone.)

99x = 4182

x=4182/99 = 1394/33.

57.357357 has 3 repeating digits so n=3.

x=57.357357 and 10^3 times that is 1000x=57357.357357

999x = 57300 ---> x=57300/99=19100/33

I have no idea how to write out the binary division here. If you email (send a thank you which will then send me your email address) I will send an image of that binary division solved.

You can also divide by converting to base 10, then converting the answer back to binary.

Remember, for binary the places to the left of 0 are 2^0, 2^1, 2^2, etc.

10,010,101 = (reading from the right) 1*1+0*2+1*4+0*8+1*16+0*32+0*64+1*128 = 149

1010 is (from right) 0*1+1*2+0*4+1*8 = 10.

The problem is 149/10 = 14 R 9/10.

In binary 14 is (from right) 0*1+1*2+1*4+1*8 = 1110

9 is (from right) 1*1+0*2+0*4+1*8 = 1001 and 10 is (from right) 0*1+1*2+0*4+1*8=1010.

The answer is 1110 R 1001/1010

P=2L+2W.

If P is even then L and W could be both odd, both even, or one odd and one even.

P = 26 = 2L+2W and 26/2=13 = L+W so if one is even the other is odd. (12+1, or 11+2, or 10+3, or 9+4 etc.). Also note that either the length or width can be odd, and the other even.

P=32 =2L+2W and 16=L+W. Both L and W can be even, or both can be odd. (9+7=16, 10+6=16.)

We can't tell anything about L and W as far as being even or odd just by knowing the P is even.

The second case, P being odd, is not possible. P = 2L +2W. 2 times an even number is even and 2 times an odd number is even and even + even is always even.

I provide online tutoring ($30/hr) and personal problem solving
($3.50-$5.50 per problem. Contact me (check the profile "website" which is
my email address.

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It's 1426, not an estimate.

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4.2-3 = 4.2-3.0 = 1.2


So 4.2-3 = 1.2

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------------
I wouldn't call that solving.

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B) Randy Johnson had an ERA of 2.64 in 2000.
At this rate, how many runs would he be expected to allow in 100 innings pitched?
:
let r = no. of runs
:
The formula: Runs, divided by innings pitched, times 9 = ERA
:
* 9 = 2.64
multiply by 100, get rid of the fraction, results
r * 9 = 100(2.64)
9r = 264
r =
r = 29 innings
:
:
Check this by finding the ERA with 29.33 innnings
* 9 = 2.64

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You give us a problem and show what you have done so far, then we help you solve it.

John

My calculator said it, I believe it, that settles it

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E=MC²

the squared character (superscript 2) is generated by holding down the alt-key
and entering 253 on the numeric keypad
___ 253 is the decimal number for the ascii code of the character

it is usually easier to utilize the caret symbol (^) to denote exponents
___ the caret is above the 6 on the keyboard

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27,900

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X-present age of the woman
Y-present age of the daughter
X+Y=46 => X=46-Y (1)
(X+4)/(Y+4)=7/2 => 2(X+4)=7(Y+4) => 2X+8=7Y+28 => X=(7Y+20)/2 (2)
NOW (1)=(2):
46-Y=(7Y+20)/2
92-2Y=7Y+20
72=9Y
Y=8
X=46-8=38

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They're equal.

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= 0.0757

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is .080 >.067?
-----
Yes because 80 is greater than 67.
Cheers,
Stan H.

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24/10
is the same as 24 divided by 10
2 4/10
reduces to
2 2/5

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= %

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50.2-600 how do I solve this problem
:
Find the difference between the larger number and the smaller number
Apply the sign (+or-) of the larger number (-600)
600.0
50.2
------
549.8, apply the sign of -600, -549.8 is the answer

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84.48/0.88=96
What this question is saying is 84.48 is 88% of what number.