Sunday

March 1, 2015

March 1, 2015

Total # Posts: 29,207

**Geometry**

did that already got this far Join AC let AC = y in triangle ADC y^2 = 22^2 + 30^2 -2(22)(30)cos D = 1384 - 1320cosD in triangle BAC, let AB = a y^2 = a^2 + a^2 - 2(a)(a)cos (2x+34°) = 2a^2 - 2a^2 cos (2x+34) so now set 2a^2 - 2a^2 cos (2x+34) = 1384 - 1320cosD Was there ...
*January 14, 2015*

**MATH!!!! AHHH!!!**

No, you want a "system", which means you need two equations. My new system was: 11x + 7y = 13 and 6x + 4y = 8 The original 8x + 5y = 9 was replaced by the sum of the two original equations, and the original 3x + 2y = 4 was replaced by a multiple of that (I multiplied...
*January 14, 2015*

**MATH!!!! AHHH!!!**

A: let's add them as suggested: 11x + 7y = 13 multiply the 2nd by 2 6x + 4y = 8 Note: the system : 11x+7y = 13 and 6x + 4y = 8 is not unique, I could have done anything I wanted here. B. ok, let's solve the original system: #1 times 2: 16x + 10y = 18 #2 times 5: 15x + ...
*January 14, 2015*

**Math**

yes, d) is also correct
*January 14, 2015*

**found it - Math**

a) correct b) tickets -- cost 20 ----- 20(12.5) or 250 , that was given 32 ----- 32(12.5) or 400 36 ----- 36(12.5) or .... 40 ----- ........ c) y(12.5) + x(12.5) or
*January 14, 2015*

**Arggghhh -Math**

Looks like you double-posted this, and one of the monitors deleted one of them. And I had just answered the one that got deleted.
*January 14, 2015*

**Algebra 1**

Come on Will Surely you can do this one!
*January 14, 2015*

**Algebra 1**

Meg --- m Jill --- m-2 Beth = (1/2)(m) = m/2 or = (1/2)m other possibly: let Jill be x Meg --- x+2 Beth --- (x+2)/2 or = x/2 + 1 or = (1/2)x + 1 try it by letting Meg = y
*January 14, 2015*

**Math**

Gab --- x Fel --- x+6 Mom --- 2(x+6) = 2x + 12 aunt --- (2x+12) + 6 = 2x + 18
*January 14, 2015*

**math**

If I read your question correctly 3/5 : 3/8 ----> divide each term of the ratio by 3 = 1/5 : 1/8 ---> multiply each term of the ratio by 40 = 8 : 5 ----> divide each term of the ratio by 1 = 8/5 : 1 I don't get 3/8 : 1 either.
*January 14, 2015*

**math**

number of cases, not worrying about order: 2hops 1 hop 6 0 5 2 4 4 3 6 2 8 1 10 0 12 number of ways each of the above can be arranged e.g. the 5 double hops and 2 single hops could be dddddss sddddds etc 6 0 --- 1 5 2 -- 7!/(5!2!) = 21 4 4 -- 8!/(4!4!) = 70 3 6 -- 9!/(3!6!) = ...
*January 14, 2015*

**Algebra- Middle**

The question makes no sense. Equations do not have a numerical value. They should have asked Is the solution to 1/3(x-9) = 2x+7 greater than or less than the solution to 4/3(x+4) = -4x To say that one equation is greater than another equation makes absolutely no sense. As to ...
*January 13, 2015*

**math**

0 = -16t^2 + 250 16t^2 = 250 t^2 = 250/16 t = √250/4 = appr 3.95 or 4.0 sec to the nearest tenth
*January 13, 2015*

**algebra**

Quick mental calculation for you k = 20 LS = -10-40 = -50 RS = -2 - 16 = -18 nope, it is wrong. Now show me what steps you had, so I can help you.
*January 13, 2015*

**calc**

then a|1 0| |0 1| + b|1 0| |0 0| = |6 0| |0 5| hope this lines up .... so |a 0| |0 a| + |b 0| |0 0 | = |6 0| |0 5| only true if: a+b = 6 a+0+5 a = 5 , b = 1 makes it true yes, it can be expressed as a linear combination
*January 13, 2015*

**math**

slope of given line is 5 so the slope of the perpendicular line has to be -1/5 so it could only be B) or C) but (-10,3) satisfies only equation C) so C
*January 13, 2015*

**Math**

If by side unit, you mean "slant height" then the area of the cone = πrl, where l is the slant height, so Area = π(15)(28) = appr 1319
*January 13, 2015*

**math**

time at slower speed --- x hrs time at faster speed ---- 4-x hrs 54x + 60(4-x) = 225 solve for x, then evaluate 54x to find distance at slower speed
*January 13, 2015*

**Math**

not enough data
*January 13, 2015*

**Algebra 2**

confirm that you mean: (a−1)+a^2−b^21−ab+2(a−b)^(2b)
*January 13, 2015*

**math**

third part --- x other part --- x+5 each solve: x + x+5 + x+5 = 31
*January 13, 2015*

**math**

ummmmhh, On the first flip, it will land either heads or tails or an extremely low probability that it will land on its edge. Would you believe that they actually did an experiment at Harvard for that condition http://adsabs.harvard.edu/abs/1993PhRvE..48.2547M Using their ...
*January 13, 2015*

**math**

Well, we don't like it I just makes things confusing in answering questions.
*January 13, 2015*

**math**

http://www.jiskha.com/display.cgi?id=1421157183 Why do you keep switching names ?
*January 13, 2015*

**math**

complementary angles add up to 90 so if your angle is x its compliment would be 90-x twice its complement would be 2(90-x) "angle if it is 6 degree less than twice its complement" ---> x = 2(90-x) - 6 solve for x, then check if your answer satisfies the "...
*January 13, 2015*

**math**

not very likely! that would make the perimeter of the square alone equal to 512 wouldn't your equation have been 4x + 3(x-5) = 251 ?? I get x = 38 (which meets all the conditions)
*January 13, 2015*

**math**

I will set up your definitions let each side of the square be x then each side of the triangle is x-5 "the sum of the perimeters of the two figures is 251 cm" translate this into your equation, then solve for x Replace you x value in the definitions. let me know what...
*January 13, 2015*

**math**

width --- x length --- x+12 perimeter = 2xwidth + 2xlength 2x + 2(x+12) = 60 etc should be easy
*January 13, 2015*

**math**

short piece ---- x longer piece ---- 2x x + 2x = 34 take over
*January 13, 2015*

**trigonometry**

Did you make a sketch ? I filled in all your data , and you actually have redundant information we can simply take tan 39° = h/3 h = 3tan39 = 2.44 m
*January 13, 2015*

**math,trigonometry**

Write down the formula, then read it out loud
*January 13, 2015*

**math**

prob = (1/12)(1/7) = 1/84
*January 13, 2015*

**algebra**

1. No, she picks only one clip If we consider only the clips, there are 12 clips that are either blue or yellow, so the prob(blue or yellow) = 12/24 = 1/2 I have ignored the pushpins, since I assume she is able to distinguish between paper clips and pushpins when randomly ...
*January 13, 2015*

**Algebra**

You could have easily checked your answer by replacing them in the original question. here is what I did: 7c + 6o = 366 , I will work in cents 10c + 3o = 339 double the 2nd: 20c + 6o = 678 7c + 6o = 366 subtract them: 13c = 312 c = 24 back into 1st: 7c + 6o = 366 7(24) + 6o = ...
*January 13, 2015*

**math**

NOW: Martha --- x John ----- x+2 After 10 years: Marta ---- x+10 John ------x + 12 their sum 10 yrs from now = 2x + 22
*January 12, 2015*

**propability**

Since he puts the olives back, the second result is independent of the first prob(black,green) = (15/36)(21/36) = 35/144
*January 12, 2015*

**math**

women -- 3x men ---- 3x/2 children -- 9x men + children = 3x/2 + 9x = 3x/2 + 18x/2 = 21x/2
*January 12, 2015*

**math**

total coins ---- x foreign coins --- x/4 mexican coins -- (2/5)(x/4) = x/10 non-mexican foreign coins = (3/5)(x/4) = 3x/20 so 3/20 of the collection is non-Mexican foreign coins check: assume there are 60 coins so 1/4 are foreign or 15 of them 2/5 of those are Mexican or 6 are...
*January 12, 2015*

**math**

The logic of this question escapes me. Why would a "preferred" CD earning 1% have an advantage over a Standard CD that earns 2% anyway, so suppose he invests his $8000 at 2% he wants $2000 interest (to get to the 10,000 minimum ?? ) 2000 = 8000(.02)t t = 2000/(8000(....
*January 12, 2015*

**algebra**

for the first one, 5n - 3 = 447 5n = 450 n = 90 Yes, 447 is in the sequence and it is term(90) D) 14-3n = 447 -3n = 433 n = -433/3 , makes no sense, since n must be a natural number So, 447 cannot be a term in this sequence Do the others the same way
*January 12, 2015*

**Math**

jog : walk = 1/2 : 3/4 = 2 : 3 = 2/3 : 1 he jogs 2/3 mile for every mile he walks
*January 12, 2015*

**Calculus**

if f ' (x) > 0 , then the function is increasing if f ' (x) < 0 , then the function is decreasing so find the first derivative, and investigate it
*January 12, 2015*

**Math**

since the average of the 5 flight is 2 : 28 the total time was 5(2:28) = 10:140 = 12:20 first 4 flights: 2 : 10 3 : 20 1 : 15 2 : 40 ------ 8 : 85 = 9 : 25 last leg = 12:20 - 9:25 = 2 : 55 Too bad the world can't can convert to "metric time" It would make things ...
*January 12, 2015*

**Geometry**

Steve is right, I can't be bothered to find my error in all my mess.
*January 12, 2015*

**Geometry**

let the radius of circle R be r let the radius of circle W be w recall that arclength = rØ, where Ø is the angle in radians arc length on first circle = (π/2)r arc length of 2nd circle = (π/3)w but they are equal πr/2 = πw/3 r = 3w/2 area of ...
*January 12, 2015*

**Math**

yes
*January 12, 2015*

**Calculus**

yes, start with a = -5 v = -5t + c , where c is a constant given: when t = 0 , v = 25 25 = -5(0) + c c = 25 ,so v = -5t + 25 when car stops, v = 0 0 = -5t + 25 t = 5 it takes 5 seconds to stop
*January 12, 2015*

**12 U Math**

So I read it as: log2 x + log2(x+2) = 3 log2(x(x+2)) = 3 x(x+2) = 2^3 x^2 + 2x - 8 = 0 (x+4)(x-2) = 0 x = -4 OR x = 2 but in log x, x≥0 by definition x = 2
*January 12, 2015*

**Algebra**

no, they are perpendicular I one line has a slope of a/b then a perpendicular line has slope of -b/a , that is, one slope is the opposite reciprocal of the other. That is the case here: 3/2 vs -2/3
*January 12, 2015*

**Algebra**

slope of first line = 3/2 slope of 2nd line = -2/3 draw your conclusion based on what you must have learned about lines and their slopes
*January 12, 2015*

**Maths Mechanics**

So make sketch , on the y-axis mark off 300 From there draw a 45° angle line ( NW) and make it 40 units Joint the endpoint to the origin. That last line is your resultant I see a cosine law application Two sides of 300 and 40 with an angle of 135° between them
*January 12, 2015*

**Algebra**

a line parallel to y = 3x - 5 must look like y = 3x + b, that is, it can only differ in the constant, the slope stays the same. So we need to find that constant b. But we know that (3,1) lies on it, so in y = 3x + b 1 = 3(3) + b b = -8 new equation: y = 3x - 8
*January 12, 2015*

**math**

n + 20 = 80 , becomes n + 3 = ??? what did you do to 20 to bring it down to 20 ?? I bet you subtracted 17. So do the same thing to the right side. You said "I know whatever I do on one side I do on the other side " , but you didn't do it
*January 12, 2015*

**Math**

-2x^2+8x-6>0 2x^2 - 8x + 6 < 0 x^2 - 4x + 3 < 0 (x-3)(x-1) < 0 y = x^2 - 4x + 3 is a parabola opening upwards with x-intercepts at 3 and 1 so it is below the x-axis, ( < ) , for 1 < x < 3
*January 12, 2015*

**Math**

5^(4/3) = (5^4)^(1/3) = cuberoot (625) I would consider 5^(4/3) as simplified
*January 12, 2015*

**Math**

mmmh, "square root" always means index 1/2 "square" always means index 2
*January 12, 2015*

**Math**

I have no clue what you mean. Type it out in symbols instead of words To me "2square root 6 the same as square root 6" means 2√6 = √6
*January 12, 2015*

**Math**

Is 2√6 = √6 ??????? Are 2 apples the same as 1 apple ? Are 200 kg equal to 100 kg ? etc ? Don't you have a calculator?
*January 12, 2015*

**Math**

mmmmh, can you solve 25t - 80 = 60 ?
*January 12, 2015*

**Algebra**

children --- x adults -----9-x 210x + 315(9-x) = 2205 solve for x , should be straightforward. BTW, $315 for a trip to Hawaii? Time to update that Math textbook.
*January 12, 2015*

**Math**

ln (x^3) is indeed equal to 3ln x perhaps your calculator is reading your input as (ln x)^3 which of course is different verification by Wolfram: http://www.wolframalpha.com/input/?i=plot+y+%3D+log%28x%5E3%29+%2C+y+%3D+3log%28x%29 now change your input line from ln(x^3) to ln(...
*January 12, 2015*

**Math**

unless your ( bracket has a different meaning from the [ bracket, they are the same, both equal +9 or plain ol' 9
*January 12, 2015*

**addmath A.P.**

if in an AP, 2x+y - x = 9x-y - (2x+y) x + y = 7x - 2y 3y = 6x y = 2x also: x + 2x+y + 9x - y = 60 12x = 60 x = 5 , then y = 10 so the first term is x or 5 common difference = 2x+y - x = x+y = 15 term11 = a+10d = 5 +150 = 155 check: first 3 terms are: 5 , 20 , 35, ... is their ...
*January 12, 2015*

**addmath**

first condition: 3k - k = k+1 - 3k 4k = 1 k = 1/4, you had that so we have *, *, *, *, k , 3k, k+1 a+5d = 3k = 3/4 a + 4d = k = 1/4 subtract them d = 1/2 in a+4d = 1/4 a + 2 = 1/4 a = 1/4-2 = -7/4
*January 12, 2015*

**Pre-Cal**

let the radius of the circle be r let the side of the square be x 4x + 2πr = 360 2x + πr = 180 or x = (180- πr)/2 x^2 = πr^2 x = r√π (180 - πr)/2 = r√π 180 - πr = 2r√π 180 = 2r√π + πr r(2√&#...
*January 12, 2015*

**trigo**

in your sketch mark the top of the mountain as P and its base as Q Label AB = 562 and mark you angles In triangle ABP angle A = 22° 30' angle B = 180° - 30° 40 = 149° 20' angle C = 180 - A-B = 8° 10' by the Sine Law, you can find BP , which is ...
*January 12, 2015*

**trigonometry**

There are two different cases here, you don't say if they are on the same side of the plane, or opposite sides of the plane. (eg. both looking east , or one looking east the other looking west) I have them on opposite sides, horizontal distance from plane of #1 tan22.5 = ...
*January 12, 2015*

**Please check my Calculus**

All 3 are correct
*January 12, 2015*

**alegbra 2**

((x+h)^(-3)-x^(-3))/(h) = ( 1/(x^3 + 3x^2h + 3xh^2 + h^3) - 1/x^3) )/h common denominator at the top = ( (x^3 - x^3 - 3x^2h - 3xh^2 - h^3)/(x^3(x^3 + 3x^2h + 3xh^2 + h^3) )/h = ( -3x^2 - 3xh - h^2)/(x^3(x^3 + 3x^2h + 3xh^2 + h^3) ) checking: if I let h = 0 this reduces to -3x^...
*January 12, 2015*

**Calculus**

the top factors, so f(x) = (x+1)(x+3)(x+4)/(x+4) = x^2 + 4x + 3 , x ≠ -4 f ' (x) = 2x + 4 = 0 for a max/min 2x + 4 = 0 x = -2 Since the function is basically the parabola y = x^2 + 4x + 3 , with a hole at (-4,3) which opens up, x = -2 will produce a minimum f(-2) = 4...
*January 11, 2015*

**Calculus**

f ' (x) = 15x^4 - 15x^2 = 0 for a max/min 15x^2(x^2 - 1) = 0 x=0 or x=±1 to determine if these yield a max or a min look at the 2nd derivative f '' (x) = 60x^3 - 30x f "(0) = 0 , so x = 0 yields a point of inflection f "(1) = 60-30 > 0 , so x = ...
*January 11, 2015*

**Reiny**

Well Dustin, after teaching this stuff for about 35 years, it becomes part of your "permanent memory". I have been retired for 17 years and I try to keep my mind active doing this voluntary tutoring. I taught in Ontario, all grade levels and all math courses, but ...
*January 11, 2015*

**math**

Derek , how are you
*January 11, 2015*

**derek --> steven ? math**

why are you switching names
*January 11, 2015*

**Geometry**

why not use the value of π on you calculator I get r^2 = 41.730426.. r = 6.4599
*January 11, 2015*

**Algebra 2**

a + k = 240 5a + 3.5k = 1140 solve them
*January 11, 2015*

**Algebra 2**

any equation of the form y = k is a horizontal line any equation of the form x = c is a vertical line so from the point (2,3) x = 2
*January 11, 2015*

**college pre-calculus**

hint: complete the square
*January 11, 2015*

**college pre-calculus**

writing down the equation by skipping the equal signs is not "showing some work"
*January 11, 2015*

**college pre-calculus**

show me what you have done so far.
*January 11, 2015*

**ap college pre-calculus**

3|x+4| = 8 means: 3(x+4) = 8 OR 3(-x-4) = 8 solve each of those Derek, you are showing no effort or steps in any of these problems. Do you simply want us to do your homework or assignment ??? I have answered quite a few of your problems. The level of these is quite simple, and...
*January 11, 2015*

**college pre-calculus**

again, your lack of brackets make the question ambiguous. The way you typed it ... why would you write y/1 that would simply be y so you really have: (y+1)(y - y^2) = y^2 - y^3 + y - y^2 = -y^3 + y if you meant: (y+1)(y/(1-y^2) = y(y+1)/((1-y)(1+y) ) = y/(1-y) , y ≠ -1
*January 11, 2015*

**college pre-calculus**

√(4/3) - √(3/4) = 2/√3 - √3/2 LCD = 2√3 = (4 - 3)/(2√3) = 1/(2√3) let's rationalize: 1/(2√3) * √3/√3 = √3/6 looks like c)
*January 11, 2015*

**college pre-calculus help**

good grief! How can anybody tell what this means with all those /'s You will definitely need brackets to establish the order of operation.
*January 11, 2015*

**Algebra Help!!**

wow, it can't get any easier just add them: 5y = 20 y = 4 sub into the first: x + 12 = 20 x = 8 x = 8, y = 4
*January 11, 2015*

**trigonometry**

cotØ + tanØ = csc^2 Ø + sec^2 Ø in your identity repertoire you should have the following: sec^2 Ø = 1 + tan^2 Ø csc^2 Ø = cot^2 Ø + 1 cotØ + tanØ = 1 + tan^ Ø + cot^ Ø + 1 0 = 2 which is false...
*January 11, 2015*

**Algebra Help!!**

follow the steps I showed you a few posts back
*January 11, 2015*

**Algebra help please!**

y = 2-x or x+y = 2 2x+2y = 4 x+y = 2 the two equations are the same so there is an infinite number of solutions.
*January 11, 2015*

**Algebra 1**

sub the 1st into the 2nd (8x-9) - 8x = -4 -9 = -4 well, that is silly! ahh, we must have 2 parallel lines no solution!
*January 11, 2015*

**math**

#1 slope = (5-4)/(-2-1) = - 1/3 , so c)
*January 11, 2015*

**math**

#2, b) would give you a slope of 5/3 so you need d) #3. (y+1)/(-4) = 1/4 4y + 4 = -4 4y = -8 y = -2 , so b) 4. the slope of any vertical line is undefined (the slope of a horizontal line is 0)
*January 11, 2015*

**Math**

Notice that we can write 4^(2x) as (4^x)^2 and you equation becomes a quadratic let 4^x = k then your equation becomes: k^2 - k - 20 = 0 (k-5)(k+4) = 0 k = 5 or k = -4 then 4^x = 5 log 4^x = log 5 x log4 = log 5 x = log5/log4 = appr 1.161 and 4^x = -4 x log4 = log (-4), but ...
*January 11, 2015*

**12 Math**

I will assume you meant 3^(x-2) = 5^x or else we would have ourselves a real mess take log of both sides and use rules of logs (x-2)log3 = x log5 x log3 - 2log3 = x log5 x(log3 - log5) = 2log3 x = 2log3/(log3-log5) = appr -4.30 check: LS = 3^(-4.3 -2) = .000985 RS = 5^(-4.3...
*January 11, 2015*

**Algebra 2**

How about just adding them as they sit 3x^2 = 69 x^2 = 23 x = ±√23 plug x^2 = 23 back into the 2nd 23 + 5y^2 = 68 5y^2 = 45 y^2 = 9 y = ±3 so 4 solutions: (√23,3) , (√23, -3) , (-√23 , 3) , and (-√23 , -3)
*January 11, 2015*

**Algebra 2**

You should be familiar with the vertex form of a parabola y =a(x-p)^2 + q , with vertex (p,q) so for yours, vertex is (-4,2) , so y = a(x+4)^2 + 2 sub in (-3.0) 0 = a(1)^2 + 2 a = -2 y = -2(x+4)^2 + 2
*January 11, 2015*

**AP Calc**

I did this for you a few days back, when you posted it as Lynds http://www.jiskha.com/display.cgi?id=1420764439 and I just noticed an error with signs I had: now dy/dx = ( (e^2x + 1)(2e^2x) - (e^2x - 1)(2e^2x) )/(e^2x + 1)^2 = (2e^4x + 2e^2x - 2e^2x + 2e^4x)/(2e^2x + 1)^2 = 4e...
*January 11, 2015*

**Math**

welcome
*January 11, 2015*

**Math**

The have given you everything you have to know in the description. so since V = lxwxh = (100-2x)(100-2x)(x) = x(100-2x)^2 SA = base + 4 sides = (100-2x)^2 + 4x(100-2x) = (100-2x)[ 100-2x + 4x] = (100-2x)(100+2x) ratio of V/SA = x(100-2x)/( (100-2x)(100+2x) ) = x/(100+2x) , x...
*January 11, 2015*

**mathematics**

"the sum of the 6th and 7th terms of an arithmetic progression is 60" --> a+5d + a+6d = 60 2a + 11d = 60 , #1 " the 3rd term is -5 " ---> a+2d = -5 , #2 Solve the two equations in two unknowns using your usual method. I would double the 2nd and ...
*January 11, 2015*

**Calculus**

have some fun at Wolfram . Change the value of the constant to see what effect it has on your graph. http://www.wolframalpha.com/input/?i=plot+y+%3D+%3Dx%5E2%2F%28x%5E2%2B5%29+ It appears as if the shape does not change, but if you look at the scale , what Wolfram is doing is ...
*January 11, 2015*

Pages: <<Prev | 1 | 2 | 3 | 4 | 5 | 6 | 7 | **8** | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>