New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'Earlsdon'
Earlsdon answered: 6288 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299, >>NextProofs/97605: I want to know the next four letters of this pattern: T,F,S,N,E,T 1 solutions
Answer 71002 by Earlsdon(6294) on 2007-09-11 21:01:38 (Show Source):
You can put this solution on YOUR website!Here's a guess!
T, F, S, N, E. T, F, S, N, T,
What do they mean...my guess is they are the initial letters of the odd integers beginning with 3.
T = 3
F = 5
S = 7
N = 9
E = 11
T = 13
F = 15
S = 17
N = 19
T = 21
|
Finance/97502: This question is from textbook
Heres the question:
A farmer wants to buy 100 animals for $100.00 (must buy at least one of them)
Horses-$5.00
Cows-$3.00
Chicks-.50
How many horses,cows and chicks does the farmer need to have 100 and 100.00? 1 solutions
Answer 70951 by Earlsdon(6294) on 2007-09-11 18:28:20 (Show Source):
You can put this solution on YOUR website!This problem involves diophantine equations, where you have more unknowns than equations.
The solutions, however, are limited by the fact that the unknowns must be integers, i.e. whole cows, whole horses, etc.
We can set up two equations as follows: H = number of horses, C = number of cows, and ch = number of chicks.
From the problem description, we can write:
H+C+ch = 100 The total number of animals is 100.
($5)H+($3)C+($0.50)ch = $100 The total amount the farmer spends.
The two equations are:
1) H+C+ch = 100
2) 5H+3C+0.5ch = 100 Multiply this equation by 2 and subtract equation 1) from the result.
1) H+C+ch = 100
2a) 10H+6C+ch = 200
--------------------
3) 9H+5C = 100 Subtract 9H from both sides.
3a) 5C = 100-9H Now divide both sides by 5.
3b) C = 20-(9/5)H Now C can only be an integer if H is a multiple of 5. Let's try H = 5.
3c) C = 20-(9/5)(5)
C = 20 - 9
C = 11 This is one possibility. Now let's try H = 10
3d) C = 20 - (9/5)(10)
C = 20 - 18
C = 2 This is another possibility.
H cannot be greater than 10 otherwise you would end up with C being a non-integer or a negative number.
So, if C = 11 when H = 5, then ch = 100-11-5 = 84 - the number of chicks.
Let's check this solution.
H+C+ch = 100
5+11+84 = 100 and...
5($5) + 11($3) + 84($0.50) = $25 + $33 + $42 = $100 This is a valid solution!
Now if C = 2 when H = 10, then ch = 100-2-10 = 88 - the number of chicks.
Let's check this solution.
H+C+ch = 100
10+2+88 = 100 and...
10($5)+2($3)+88($0.50) = $50+$6+$44 = $100 This solution is also valid!
The two solutions are:
1) 5 horses, 11 cows, and 84 chicks.
2) 10 horses, 2 cows, and 88 chicks.
|
Miscellaneous_Word_Problems/97428: iN YOUR ECONOMICS CLASS YOU HAVE SCORES OF 68, 82, 87, AND 89 ON THE FIRST FOUR OF FIVE TEST. TO GET A GRADE OF B THE AVERQAGE OF THE FIRST FIVE TEST SCORES MUST BE GREATER THAN OR EQUAL TO 80 AND LESS THAN 90. SOLVE AN INEQUALITY TO FIND THE RANGE OF THE SCORES THAT YOU NEED ON THE TEST TO GET A B 1 solutions
Answer 70905 by Earlsdon(6294) on 2007-09-11 11:40:55 (Show Source):
You can put this solution on YOUR website!To find the average score (A) of the first five tests, you'll add the scores and divide by 5.
 where x is the score on the fifth test.
Now you need this average (A) to be at least 80 (  ) but less than 90 (  ).
You can write the inequality thus:
 To solve this, you'll simplify the numerator.
 Now you'll multiply everything by 5.
 Simplify.
 Now you'll subtract 326 from everything.
 Simplify.
 Read this as: x is at least 74 but less than 124.
The range of scores needed to get a B average is at least 74 but less than 124.
|
Equations/97151: Find two consecutive odd integers such that 3 times the first integer
is 5 more than twice the second. Please help me figure this one out.
I also would like to no if there is a way that I can check to see if the question I am about to ask is not already been answered somewhere? Can I search the web site some how? Could you tell me how if it is possible?
Thank you 1 solutions
Answer 70728 by Earlsdon(6294) on 2007-09-09 22:14:44 (Show Source):
You can put this solution on YOUR website!1) Let x be the first odd integer, then x+2 will be the next consecutive odd integer.
From the problem description, you can write:
3x = 2(x+2)+5 Apply the distributive property to the right side.
3x = 2x+4+5 Simplify.
3x = 2x+9 Subtract 2x from both sides.
x = 9
The two integers are 9 and 11
Check:
3(9) = 2(11)+5
27 = 22+5
27 = 27
2) I don't know of an easy way to see if a similar problem to one you want help with has already been answered. You could try clicking on "Recent problems solved by all" and do a manual search but that could be quite time-consuming.
|
Travel_Word_Problems/96972: A bike race consists of two segments whose total length is 90 km. The first segment is covered at 10 kph and takes 2 hours longer to complete than the second segment, which is covered at 25 kph. How long is each segment?
I have tried setting it up as t/25 = (t+2)/10 but for some reason I can't figure out the algebra.
Thanks! 1 solutions
Answer 70600 by Earlsdon(6294) on 2007-09-09 11:41:04 (Show Source):
You can put this solution on YOUR website!From your equation, it appears that you assumed that the two segments were the same distance, but there is nothing in the problem to support this assumption.
So, back to basics:
 and...

We know that:
 and...
 and we also know that:
 So we can write:
 aagh...to many variables! Let's replace  with  and replace  with  , so now we have:
 and
 Solve this for
 ...but, of course,  , so...
 Good, now we have one variable  , so let's solve for it.
 and...
 hrs. Now we can find  and
 km and...
 km
Check:
Total distance is  = 90km
|
Exponents/96847: how do you answer this question? (2^4)(2^3) I tink the answer may be negative exponent law = 1/2^1 1 solutions
Answer 70470 by Earlsdon(6294) on 2007-09-08 11:52:22 (Show Source):
You can put this solution on YOUR website!Simplify:
 When multiplying quantities with exponents, if the bases are the same (base is 2 in this case) then you add the exponents:
 =
|
Finance/96338: This question is from textbook prentice hall mathematics algebra 1
three friends were born in consecutive years. the sum of their birth years is 5961. find the year in which each person was born.
i need help solving this problem. 1 solutions
Answer 70126 by Earlsdon(6294) on 2007-09-05 18:14:41 (Show Source):
You can put this solution on YOUR website!Let x = the year of birth of the youngest, then x+1 will be the year of birth of the next oldest and x+2 will be the year of birth of the eldest friend.
So we need to find x.
x+(x+1)+(x+2) = 5961
3x+3 = 5961 Subtract 3 from both sides.
3x = 5958 Divide both sides by 3.
x = 1986
x+1 = 1987
x+2 = 1988
Check:
1986+1987+1988 = 5961
|
Linear-systems/96328: This question is from textbook
You travel 575 miles in 8 hours. Part of the trip is made at 60mph and part of the trip is made at 80mph. Find the time traveled at each rate. This is the question. I have that distance =rate * time and 575=r*8. I also have that r=(60T+80t)/2 and T+t=8 but when i try to solve it I get a negative time. Thanks for trying to help. 1 solutions
Answer 70122 by Earlsdon(6294) on 2007-09-05 17:58:56 (Show Source):
|
percentage/96265: If you have a 5% solution and a 50 % solution how much of each solution do you need to make 20ml of 30% solution? 1 solutions
Answer 70083 by Earlsdon(6294) on 2007-09-05 14:43:19 (Show Source):
You can put this solution on YOUR website!Let x = the number of ml of the 5% solution required, then 20-x = the number of ml of the 50% solution needed. The sum of these will equal 20 ml of 30% solution.
From these facts, we can write the necessary equation to find x, after converting the percentages to their decimal equivalents.
x(0.05)+(20-x)(0.5) = 20(0.3) Simplify and solve for x.
0.05x+10-0.5x = 6
-0.45x+10 = 6
-0.45x = -4
x = 8.89
So you would need 8.89 ml of the 5% solution and 20-x = 11.11 ml of the 50% solution to obtain 20 ml of 30% solution.
|
Numeric_Fractions/96233: Please help me. I have 3 parts ti this question. I have to anwer any 2 of them and cannot get the anse=wer. This is my first time doing Intermediate Algebra. Thank you for your help.
1. Find an example from your line of work or daily life that can be expressed as a polynomial
2. What does FOIL stand for and what is the purpose of FOIL?
3. What happens to the inner and outer products in the product of a sum and difference, (a + b)(a – b)? Give an example.
1 solutions
Answer 70055 by Earlsdon(6294) on 2007-09-05 10:44:27 (Show Source):
You can put this solution on YOUR website!1) I think that you will have to find your own answer to this question but one example might be:
When I hit a baseball into the air, the height, in feet, it reaches can be expressed by the polynomial:
2) FOIL stands for:
F = First terms.
O = Outer terms.
I = Inner terms.
L = Last terms.
The purpose of FOIL is a mnemonic (memory aid) which reminds you of the order in which to multiply the terms of two binomials.
For example, to multiply the binomials:
 you would multiply the first term (F) of each binomial ((2x)(3x)), then the two outer terms ((2x)(4)), then the two inner terms ((3)(3x)), and finally the two last terms ((3)(4)), so it looks like this:
 then you would simplify this by combining like-terms to get:  as the answer.
3) The inner and outer products of the product of the sum and difference will add up to zero:
 and when you combine like-terms, the +ab and -ab add up to zero so you are left with:
|
Sequences-and-series/96045: I want to neqotiate an allowance with my parents propose that I make 1 cent the first week 2 cents the 2nd week 4 cents the 3rd week and 8 cents the 4th week, etc how much will I make after 52 weeks which happens to be total weeks for a year How much will I make the last week 1 solutions
Answer 69948 by Earlsdon(6294) on 2007-09-04 16:05:50 (Show Source):
You can put this solution on YOUR website!This is a case of a geometrical progression and can be repesented as...
 +...+ 
The sum of this progression is  cents
So, at the end of the year (52 weeks) you would have a total of:
$45,035,996,273,704.95
On the 52nd week, you would get:
 cents, which amounts to $22,517,998,136,852.48
|
Graphs/96028: A moving company charges a certain flat rate per day, plus a per-mile charge. It charged your brother $52.90 for a one-day move of 20 miles. Your sister used the same company and was charged $81.70 for a one-day move of 60 miles. What would you predict to pay for a one-day move of 85 miles? 1 solutions
Answer 69941 by Earlsdon(6294) on 2007-09-04 15:44:09 (Show Source):
You can put this solution on YOUR website!First, I would calculate the flat-rate cost (C) and the cost per mile (x). This can be done as follows:
Write the equation for the cost of the brother's move of $52.90 for a 20-mile move...
1)  ...and the sister's move.
2) 
Now you have two equations with two unknowns (C and x).
Rewrite these as:
1a)  and...
2a) 
But the flat rate, C, is the same in both cases, so equation 1a) is equal to 2a).
 Simplify and solve for x, the cost per mile.
Add 60x to both sides.
 Now subtract $52.90 from both sides.
 Finally, divide both sides by 40 to get x.
 So the cost per mile is 72 cents. Now that we have that, we can go back to either equation 1a) or 2a) to find the value of C, the flat-rate cost. Let's use equation 1a) and we;ll substitute x = $0.72:
 Simplify.
 ...and this ($38.50) is the flat-rate cost.
Now we can work out the cost for a one-day move of 85 miles by using 1) but substituting C = $38.50 and x = $0.72 per mile..
 = 
The cost of a one-day move of 85 miles is $99.70
|
Equations/95640: 3/7x +2 =4/7x-5
I have tried this over and over again - i keep coming up with -49, but the answer should be 7. I cannot figure it out.
I learned that you should multiply everything by the LCD to make everything a whole number first.
1 solutions
Answer 69672 by Earlsdon(6294) on 2007-09-02 11:53:42 (Show Source):
You can put this solution on YOUR website!I have worked your problem two different ways and find that the answer that you give (x = 7) is just not possible.
In the first place, to remove any ambiguity on the problem statement, please use parentheses! Why? Because...
3/7x can be interpreted to mean either:  or 
Whereas, (3/7)x can be interpreted only as
In the second place, the only way that I can get the answer that you give, (x = 7) is if the problem starts with  , so...
 Multiply through by the LCD of 7.
 Simplify.
 Add 3x to both sides.
 Add 35 to both sides.
 Finally, divide both sides by 7.
 or
Please check the original problem for a possible missing negative sign!
|
Equations/95641: I am having trouble. Can someone please help me.
Solve:
-3(x+1)=2(x-8)+3 1 solutions
Answer 69668 by Earlsdon(6294) on 2007-09-02 11:24:23 (Show Source):
You can put this solution on YOUR website!Solve for x:
-3(x+1) = 2(x-8)+3 First, apply the "distributive property" to both sides of the equation.
-3x-3 = 2x-16+3 Simplify.
-3x-3 = 2x-13 Add 3x to both sides.
-3 = 5x-13 Add 13 to both sides.
10 = 5x Finally, divide both sides by 5.
2 = x or x = 2
Check: Substitute x = 2 in the original equation.
-3(2+1) = 2(2-8)+3 Simplify.
-6-3 = 4-16+3 Evaluate.
-9 = 7-16
-9 = -9 It checks!
|
Functions/94994: Ok, My son came home from school today with Math homework. He is in Algebra 1 and they have requested he have a TI-83. I intend to buy this for him but I haven't yet and he has homework tonight. The teacher only explained how to do the problem on the calulator not on paper. I know it can be done on paper but I really do not remember how to do it. I looked in his math book and the book only goes to Chapter 12 and the worksheet is for Chapter 13. I can't figure out how to solve these problems and I'm hoping for some help. He only has to do certain problems so I'm going to send you one of the ones he doesn't have to do.
The question reads as follows:
Solve ecah equation if the domain is {-2, -1, 1, 3, 4}.
The question is as follows:
y=4 - 5x
If you could just show me how to work these I can help him with the ones he has to turn in tomorrow.
Thank you. 1 solutions
Answer 69127 by Earlsdon(6294) on 2007-08-28 22:46:45 (Show Source):
|
Equations/94750: I have an algebra question for you, to which I require assistance, as I am not quite certain what it is they want me to do. It is a written question.
An alloy contains 60% by weight of copper, the remainder being zinc. How much copper must be mixed with 50kg of this alloy to give an alloy containing 75% copper?
I have done some nuber crunching and all I got was
60% of 50kg = 30kg
and
75% of 50kg = 37.5kg
and or
60x+40y=50
75x+25y=50 /5
60x+40y=50 /4
15x+5y=10
15x+10y=12.5
15x+5y=10
anyhow
X = .5kg
and
Y = .5kg
but it just does not look right, would you please be able to show me the correct way to sort this out. It would be greatly appreciated. Thank you. 1 solutions
Answer 68992 by Earlsdon(6294) on 2007-08-27 23:13:06 (Show Source):
You can put this solution on YOUR website!First, let's concentrate on the amount of copper (Cu) in the starting alloy and the resulting alloy.
The amount of copper in the starting alloy is: 60% of 50 kg or (0.6)(50). To this we want to add x kg of Cu to make (50+x) kg of an alloy, 75% of which is Cu. So we set up the equation:
(0.6)(50) + x = (0.75)(50+x) Simplify and solve for x.
30+x = 37.5+0.75x Subtract 0.75x from both sides.
30+0.25x = 37.5 Subtract 30 from both sides.
0.25x = 7.5 Finally, divide both sides by 0.25
x = 30
You would need to add 30kg of Cu to the 50kg of the starting alloy to obtain 80kg (50kg + 30kg) of the resulting alloy containing 75% Cu.
Check:
50(0.6)+30 = (50+30)(0.75)
30+30 = (80)(0.75)
60 = 60
|
Polynomials-and-rational-expressions/94670: I know what I am supposed to do, and if I could find the LCD, I could solve the problem, but for the life of me, I am drawing a blank on how to factor (n^4 -1) 1 solutions
Answer 68949 by Earlsdon(6294) on 2007-08-27 18:20:44 (Show Source):
You can put this solution on YOUR website!Factor:
 Notice that  , so now you can write:
 The difference of two squares can be factored:  Applying this to your problem:
 but again, we have the difference of two squares in the second factor,  so...
 we would normally stop here, because the factors of  involve the imaginary number 
If you are interested though:
 but don't use this if you have not yet come to imaginary numbers.
|
Exponents/94556: This is a table so it's tricky typing out in words...
There are 4 rows and 4 columns, and I need to fill in the blanks.
The first column is labeled POWER. The first row in the first column says 2^5.
The fourth row in the first column says r^8. The second column is labeled BASE. The second row in the the second column says 3. The third column is labeled EXPONENT. The second row in the third column says 4. The fourth column is labeled STANDARD FORM. The third row in the fourth column says 125.
phew...
Okay, if you understood that, and if I'm correct, the entire first row should read: 2^5, 2, 5, 32. The second row should read: 3^4, 3, 4, 81. I have no idea about the third row. The last row should read: r^8, r, 8, r^8?
I'm really sorry that was so long and confusing, but I just can't get that third row!
1 solutions
Answer 68881 by Earlsdon(6294) on 2007-08-26 23:33:53 (Show Source):
You can put this solution on YOUR website!Here's where "a picture is worth a thousand word" n'est-ce pas?
Anyway, I get the picture! (lol)
You are correct for rows 1, 2, and 4. The third row is shown below:
 : Base = 2 Exponent = 5
 : Base = 3 Exponent = 4
 : Base = 5 Exponent = 3
 : Base = r Exponent = 8
|
Complex_Numbers/94510: I am desperate to know if I am doing this correctly. Would you please let me know how I am going? Thank you so much.
a=2, b=3, c=-4, d=-1, e=0
–
–
(2/3) (2/3) - (-4/-1) (-4/-1) (-4/-1)
2/3*2/3 – 4*4*4
(2*3) (3*2)/3*3 – 64
6*6/9 – 64
36/9 - 64
4 – 64
= - 60
1 solutions
Answer 68829 by Earlsdon(6294) on 2007-08-26 19:35:30 (Show Source):
|
Linear-equations/94429: graph the equation y = 2x - 3 on a coordinate plane 1 solutions
Answer 68756 by Earlsdon(6294) on 2007-08-26 10:27:46 (Show Source):
You can put this solution on YOUR website!Graph y = 2x-3 on a coordinate plane.
This is quite easy to do on your own coordinate graph paper. This is a linear equation (you can tell this because the highest power of the independent variable, x, is 1) and since you need only two points to define a straight line, you would just assign a value to x and solve for the corresponding value of y to give you one point (x, y) then do this again with a different value of x to get the second point.
Plot these two points on your graph paper and draw the straight through them.
Let's pick x = 0 for the first point. Plugging this into your equation, we can solve for the corresponding value of y, the dependent variable.
For x = 0
y = 2(0)-3
y = -3
The coordinates of the first point are (0, -3) and...
For x = 3
y = 2(3)-3
y = 6-3
y = 3
The coordinates of the second point are (3, 3)
The graph appears below:
|
Radicals/94371: 2/(3+sqrt5)
please help i forgot how to deal with the denominator 1 solutions
Answer 68720 by Earlsdon(6294) on 2007-08-25 18:14:06 (Show Source):
You can put this solution on YOUR website!Simplify? We don't like to see radicals in the denominator, so...
 Multiply the numerator and the denominator by conjugate of the denominator which is (  ).
 Simplify the denominator.
 Simplify some more.
 Cancel a 2 in the top & bottom.
|
Graphs/94341: I am a little confused. Can I get someone to help me please. I worked out the problem I am just not sure if it is correct.
Rewrite as a function of x.
4-x-10y=11
Thank you. 1 solutions
Answer 68666 by Earlsdon(6294) on 2007-08-25 12:51:10 (Show Source):
You can put this solution on YOUR website!Rewrite as a function of x:
 First, solve this for y.
 Divide both sides by 10.
 or
 Now, to write this as a function of x in function form, just replace the y with f(x)
|
Circles/94328: Please help me: It says to Complete the square to find the center of the radius of the given circle.Could you show me the steps in completing this.
A
B
Thank you 1 solutions
Answer 68665 by Earlsdon(6294) on 2007-08-25 12:40:03 (Show Source):
You can put this solution on YOUR website!A. Complete the square to find the centre and the radius.
 Add 3 to both sides.
 Group the like-terms together on the left side.
 Now complete the squares in the x- and y-terms. To do this, you add the square of half the x-coefficient and the square of half the y-coefficient to both sides of the equation.
For the x's this would be  and for the y's this would be  , so we get:
 Factor the trinomials and simplify.
 Now compare this with the standard form for a circle with centre at (h, k) and radius r.
 and you can see that the centre (h, k) is at (-3, 2) while the radius is  = 4.
Follow this approach on the second problem and see how you do.
If you still have problems, please re-post.
|
|