the side of a certain square measures 3 feet longer than that of another square. the sum of their ar

the side of a certain square measures 3 feet longer than that of another square. the sum of their ar

We'd need to know the rest of the question to complete the answer but I can give the first few steps and leave the final solution for you.
jim360
27 July 2012
Whoops, a bit of a typing fail there, give me a minute to write it all out.
jim360
27 July 2012
OK, so to go about solving this problem, we take the side of the smaller square and call it L. Then the length of the sides of the larger square must be L+3. That means that the area of the small square must be L^2, "L squared", and the area of the larger square is (L+3)^2, "L plus three, all squared". For the area of the larger square we now expand out the brackets (L+3)(L+3) using the usual rules [if you don't follow this next step let me know] and you get L^2 + 6*L + 9, where the * means "multiplied by". Now the sum of their areas is ... well, call it A, anyway. So we have that L^2 + L^2 + 6L + 9 = A (for area), or 2(L^2) + 6*L + 9 = A. Then we're now at the step of solving a quadratic equation, which you should be able to do if you're being asked this question. Hope that helps.
jim360
27 July 2012
the side of a certain square measures 3 feet longer than that of another square. the sum of their areas is 117 square feet. find the length of a side of the smaller square.
mindy129
27 July 2012
and one more:
mindy129
27 July 2012
The lengths of the sides of a rectangle are x and x+6. the area of the rectangle is 55 sq.ft. find the lengths of the sides.
mindy129
27 July 2012
thank you very much for the previous and i was thinking if u can give me the time you are on so when i need help i will go on at that time and so i guess we can chat a little...
mindy129
27 July 2012
You're welcome and I should be OK to do that at some point. What's your time zone? I'm usually free most of the time at the moment. Anyway, for the first question what answers did you get? For the second, the same start. A rectangle's area is given by base*height, which in this case is: x(x+6) = x^2 + 6x (expanding the brackets) = 55 (from the information given) So, putting it in the standard form for quadratic equations: x^2 + 6x - 55 = 0 Which can be solved using the usual methods, to get x = 5 . So the side are 5 ft. and 11 ft.
jim360
27 July 2012
Thank you and for additional information. I live in New York.
mindy129
28 July 2012
tell me the time your free
mindy129
28 July 2012
Ok well I'm five hours ahead of you so usually I'm free in your mornings and afternoons up until about 5pm your time. If you want to get with me tomorrow then that should be OK - I'll be around from about 10am your time.
jim360
28 July 2012
the length and width of an rectangle are 15 cm and 8 cm, respectively. find the length of the diagonal.
mindy129
29 July 2012
find the diagonal of a rectangle whose sides measures 5 yd. and 12 yd.
mindy129
29 July 2012
Both of those two problems can be solved using Pythagoras' Theorem, that in a right-angled triangle, the sum of the squares of the lengths of the two shorter sides equals the area of the square of the hypotenuse (longest side), or (calling the two shorter sides a and b, the hypotenuse h, and ^2 meaning "squared"), a^2 + b ^2 = h^2. For the problems you give, you've been told the lengths of the two shorter sides. So square both numbers, add them, and take the square root of the total.
jim360
29 July 2012
mindy129
30 July 2012
You're welcome. It can take a while to recognise which method to use to solve each problem but, once you get the hang of it, you should find questions like these much less of a problem. It's all about practice.
jim360
30 July 2012
may i ask what are consecutive odd integers?
mindy129
30 July 2012
for example: solve algebraically and check: the length of one leg of an right triangle is 8 in. the lengths of the other leg and the hypotenuse are consecutive odd integers. find the length of the hypotenuse.
mindy129
30 July 2012
Integers are just whole numbers that can be positive or negative, so 123, 6, 0 and -79 are all integers. Consecutive odd integers are just two odd integers that are next to each other. So, say, 1 and 3, or 99 and 101.
jim360
30 July 2012
Draw the graphs on the line bu choosing points. (Well you can't exactly teach me how to put the points in the graph but give me the (x,y) so there i know where to put it on): 1) 1/2x + y= 7 2)
mindy129
02 August 2012
refix #1 is 1/2 + y=7 2) y= 3x+1 3) 2x-3y= 10
mindy129
02 August 2012
draw the graph of the line by the intercepts method.(just show me how to do the intercepts method only) 1) y= -7/5x+ 6 2) 2y- x= 8
mindy129
02 August 2012
For all of these problems the approach is the same: Rearrange the equation of the line as an equation for y in terms of x, so that, say, #3 becomes: 2x - 3y = 10 (2/3)x + 10/3 = y (make sure you can get this result for yourself). Then, choose any value for x that you like. Ideally pick whole numbers such as 0, 1, 2, 3. Insert this value for x into your equation. Again, in the case of #3 we get: Let x = 1. So y = (2/3) * 1 + 10/3 = 12/3 = 4. So a point on the line is (x,y) = (1,4). Then, choose any other value for x. Ideally choose a point a reasonable distance away from the first point. There's no special reason for this, but it's easier to get a straight line right between points that are reasonably far apart. So you might pick x = 5 for the second point. This would give you two points, then draw a line between the two. For the "intercept method" The approach is slightly different. What we do is find where the line crosses the two axes, x and y. Along the y-axis (the vertical one) x is always zero. Along the x-axis y is always zero. So simply put x=0 into the equation for the line and find the value of y that fits, then do the same with y = 0. For #3 We'd find, for y = 0: 2x - 3*0 = 10. 2x = 10 x = 10/2 = 5. So that the intercept of the x-axis is at (5,0).
jim360
02 August 2012
That last message may not have formatted properly so that the equations may not be clear. Let me know if you can follow it.
jim360
02 August 2012
if i plug in 5 for the previous questions then i will get a decimal of 6.667. so how do i do that?
mindy129
02 August 2012
but i am still confuse how you reaverge the problem and how you know which # to put such as 1,2,3,or 4 and does y have to be like this (=y)
mindy129
02 August 2012
and for the first question why did u plus instead of the orginal question which is subtracting
mindy129
02 August 2012
and i am also confuse on 1/2x+ y=10
mindy129
02 August 2012
There's several questions there and it would be easier if we went through that in a private session. If you are around in your tomorrow morning (9am onwards) then contact me and I'll see what I can do. I should be around then.
jim360
02 August 2012
i don't think i can be free on 9 because i am usually free on 11 because i have piano class until that time and 1:30 i am going to summer school so maybe you can try to explain to me how you can actually reaverge 2y+x= 5 but i learned that you have to plug in 0,1,2,3 in after i reavege it. after that i know how to put it on the graph
mindy129
03 August 2012
When rearranging equations it depends on what you are wanting to make the "subject" of the equation, which is the bit that is equal to everything else. So in u = 14v - 1, say, u is the subject. For 2y + x = 5: If you want y to be the subject, first subtract x from BOTH sides. This cancels the x on the left as x - x is zero. So we are left with: 2y + x - x = 5 - x 2y = 5 - x. Now divide both sides by 2. Since 2/2 = 1, we find: 2y/2 = (5-x) /2 y = 5/2 - x/2 Note that you must divide everything on the right-hand side by two. That's made y the subject. As for being free, if you aren't on this morning that's not a problem but you might want to try to get with me at some point as you still have a lot of questions and it's harder to answer them all at once. Let me know when you are free, and if you are willing to pay for more involved help.
jim360
03 August 2012
Argh, again it's not formatted properly. In the equation lines, read: First line: 2y + x - x = 5 - x 2y = 5 - x. Should read: 2y + x - x = 5 - x 2y = 5 - x Second line: 2y/2 = (5-x) /2 y = 5/2 - x/2 Should read: 2y/2 = (5 - x) /2 y = 5/2 - x/2
jim360
03 August 2012
i hope i can buy tutor time but i don't think my parents allow me to. Something you just can tell what your parents think you know...because you are like part of them. But i don't mind if you can't explain it in words ten its totally fine.
mindy129
05 August 2012
Thank you for trying
mindy129
05 August 2012
sometimes*
mindy129
05 August 2012
Well there's no harm in asking - but obviously don't do anything they don't want you to do. If you have more short questions I'll try to answer them as and when they arise.
jim360
05 August 2012
the tens digit of the two digit number is 3 more than the units digit. the number is 6 less than seven times the sum of the digits. find the number.
mindy129
12 August 2012
That's a cute little problem that can be solved by setting the units digit equal to x. Then the tens digit is equal to x + 3, and their sum is 2x + 3. The second part of the question tells us that: 7(2x+3) - 6 = 10(x+3) + x Where the left-hand side comes from what the question tells us, and the right-hand side from applying place value to a two-digit number. Then you should rearrange that equation to find x.
jim360
12 August 2012