Monday

July 28, 2014

July 28, 2014

Total # Posts: 26,891

**ALGEBRA**

What is the x value of your vertex ?

**College alg.**

easy to graph, just grab some ordered pairs Parabola opening downwards, domain of any standard parabola is the set of real numbers, (any x you want to choose) the domain in this case : all values of y ≤ the maximum y (you will need the vertex)

**College Algebra**

Since you labeled it "college" algebra, I will assume you know Calculus s(t) = -4.9t^2 + 7.1t + 10 v(t) = -9.8t + 7.1 = 0 for a max of s(t) 9.8t = 7.1 t = 71/98 s(71/98) = -4.9*71/98)^2 + 7.1(71/98) + 10 = appr 12.58 m or 12.6 m above water (to the nearest tenth)

**pre algebra**

#1. first equation: LS = -12 RS = 6(-4) + 12 = -12 2nd equation: LS = 2(-4) - (-12) = -8 + 12 = 4 RS = 4 So, yes , it is a solution. #2. since y = x+4 and y = 3x , then 3x = x+4 2x = 4 x = 2 sub into either one of the two original equations ... if x = 2, y = 6 so the solution ...

**prealgebra**

correct, subbing in x=-1 has to give you +5 for both equations.

**math**

teachers --- t students --- s from first case: 8.5t + 2s = 29 from the 2nd case: 34t + 8.5s = 119 System of 2 equations in 2 unknowns. Just your favourite method of solving.

**Math**

Using standard alphabet, number of possibilities for the 4 numbers = 10^4 number of possibilities for the letters = 26^3 Three more numbers = 10^3 Number of possible plates = (10^4)(26^3)(10^3) = 175,760,000,000 note: this includes plates such as 0000ZZZ000

**Calculus**

I agree with your choice of b) for #1 If the slope of the tangent is zero, then the tangent must be a horizontal line and you are at either a maximum or a minimum point of the curve.

**Math**

I observed that the derivative of the inside of the bracket sits out front as a multiple of that. If that is the case, I just reverse by "thinking" if dy/dx = x^2(5x^3 + 9)^3 y = (x^2/(15x^2)) (5x^3 +9)^4 * (1/4)+ c = (1/60) (5x^3 + 9)^4 + c

**Math**

N '(t) = 2e-0.02t N = (2/-.02) e^-.02t + c N = -100 e^-.02t + c when t=0 , N = 45 45 = -100(1) + c c = 145 N = -100 e^-.02t + 145 testing: if t = 0 N = -100 e^0 +145 = 45 , good if t = 5 N = -100 e^-.1 + 145 = -90.5 + 145 = 54.5 , improving if t = 20 N = 78 , looks logical...

**Math**

if dy/dx = 6(2x-7)^5 y = (6/6) (2x-7)^6 (1/2) + c y = (1/2)(2x-7)^6 + c but (4 , 3/2) lies on it, so 3/2 = (1/2)(1)^6 + c 3/2 = 1/2 + c c = 1 f(x) = (1/2)(2x-7)^6 + 1

**MATH Algebra**

You must use brackets to establish the correct order of operation. According to your steps, the question was: (48x + 32y)/32 = 48x/32 + 32y/32 = 3x/2 + y you have the correct answer. <Also 25a + 5b + 15c/10 would I be correct to divide by 5?> yes, but again you will need...

**math**

number of miles ---- m .15m + 22.5 ≤ 50 .15m ≤ 27.5 m ≤ 183.333.. Suppose you drive 183 miles cost = 183(.15) + 22.50 = 49.95 , still good suppose you drive 184 miles cost = 184(.15) + 22.5 =50.10, oops, we are over so you can drive 183 miles and still be und...

**Calculus1**

Since we are including the waste for cutting the circles from squares, we need 2 squares and a rectangle if the radius is r, and our square is 2r by 2r, we can cut 2 circles from 2squares. Let the height be h , (all units in cm) we must have: 1000 =π r^2 h h = 1000/(π...

**math**

I will assume you are not taking Calculus, but have studied the quadratic function, and know how to complete the square y = -16(x^2 - 2x) = -16(x^2 - 2x + 1 - 1) = -16( (x-1)^2 - 1) = -16(x-1)^2 + 16 so the vertex is (1,16) suggesting that the max height is 16 after 1 second s...

**calculus**

This one is more obvious than your last one (e^-x - 1)/(e^-x + x) notice that the derivative of (e^-x + x) is (-1)e^-x + 1 = -1(e^-x - 1) you have exactly the opposite of that sitting in your numerator. Which , by observation alone, suggest a log integration since for y' =...

**math**

let the dimensions of the first plot be x by y let the dimensions of the second plot be a by b where x,y, a, and b are all integers. so we want: xy + ab = 60, and x-y = a+b There are many combinations that will work e.g. 15 2 30 3 10 30 x=15 y=2 a=3 b=10 area of first = 30, ar...

**Good one Bob - calculus**

and Bob saw an even third interpretation of your typing (e^-x - 1)/(e^-x+ x )^2 http://integrals.wolfram.com/index.jsp?expr=%28e%5E-x+-+1%29%2F%28e%5E-x%2B+x+%29%5E2&random=false

**calculus**

my interpretation: (e^-x - 1)/(e^-x+x^2) , you have mis-matched brackets even the usual reliable Wolfram had difficulties with that one http://integrals.wolfram.com/index.jsp?expr=%28e%5E-x+-+1%29%2F%28e%5E-x%2Bx%5E2%29&random=false the way you typed it, fixing the missing bra...

**Calculus**

v(0) = 35(25)^2 = 21875 v(15) = 35(10)^2 = 3500 avg rate of change for those 15 minutes = (2500 - 21875)/(15-0) = -18375/15 L/min = -1225 L/min You are correct at t=15, you are looking for the instantaneous rate of change, that is, the derivative of v(t) v ' (t) = 70(25-t)...

**math**

Dimensions of the rectangle including the sidewalk is 290 by 730 , (added 5 feet on each side) surface area of sidwwalk = 290x730 - 280x720 =10100 ft^2 volume of sidwwalk = 10100(8/12) ft^3 = 20200/3 ft^3 but we are ordering 5% extra volume ordered = 20200/3 (1.05) = 7070 cubi...

**Logarithmic Algebra**

log 100 = log(25*4) = log 25 + log 4 = log 5^2 + log 2^2 = 2log5 + 2log2 = 2c + 2a

**Algebra**

= x^9 y^36

**Math**

let the angle be x then its complement is 90-x x = 4(90-x) x = 360 - 4x 5x = 360 x = 72 so the angle is 72° and its supplement is 180-72 = 108° check: our angle = 72 its complement = 18 is the angle 4 times its complement ? is 72 = 4(18) ? yes

**Algebra**

so .. [(x^2 + 8x + 15)/(x-4)] * [(x^2 – 16) / (2x + 6)] = (x+3)(x+5)/(x-4) * (x-4)(x+4)/(2(x+3) ) = (x+5)(x+4)/2 , x ≠ -3, 4 the temptation is to anticipate asymptotes at x = -3 and x = 4 but since we obtain 0/0 for these two values in the original, but an actual va...

**math**

You are probably not getting any replies because we might all be confused by " Find the number of ordered N samples of size 3: " I have no idea what you mean. Are there different sizes of lightbulbs? Are we choosing lightbulbs? ???

**Calculus**

so you are solving 2sin 2t + cos 2t = 0 2sin 2t = -cos 2t 2sin 2t/cos 2t = -1 tan 2t = -1/2 set your calculator to RAD and find 2nd function tan (tan^-1 (+1/2) to get .46365 we know that the tangent is negative in II and IV so 2t = π - .46365 = 2.6779 t = 1.339 or 2t = 2&...

**math**

z-score for the 169 = (169-180)/8 = -1.375 x-score for the 191 = (191-180)/8 = +1.375 look up value for 1.375 and the value for -1.375 subtract the two values You should get .8309 I recommend this webpage for these type of questions. http://www.wolframalpha.com/input/?i=plot+y...

**math**

number of dimes ---- x number of quarters --- 21-x by value: 10x + 25(21-x) = 405 10x + 525 - 25x = 405 -15x = -120 x = 8 so 8 dimes and 21-8 or 13 quarters or dimes ---- d quarters ---q based on numbers: d+q = 21 based on value: 10d + 25q = 405 --- 2d + 5q = 81 double the fir...

**College Algebra**

when is 2 = 1 (1.3)^t take ln of both sides ln 2 = ln (1.3)^t ln2 = t (ln 1.3) t = ln 2/ln 1.3 = appr 2.64 yrs or 2 years and 7.7 months check suppose there were 5 million sold now, in 2.64 years, the number sold is N = 5 (1.3)^2.64 = 9.995 million or double the original 5 mil...

**Algebra**

no unique answer. any pair of factors of 15x3 + 25x^2 would do 15x^3 +25x^2 = x^2(15x + 25) or = 5x(3x^2 + 5x) or = 5(5x^3 + 5x^2) etc for any answer, the two factors would be the length and width

**9th grade math**

divide each term by 4 and move all terms to the left x^2 - 20x + 100 = 0 (x - 10)^2 = 0 x-10 = 0 x = 10

**9th grade math**

did you notice the left side is a perfect square? (x+4)^2 = 25 take √ of both sides x+4 = ± 5 x = -4 ± 5 = 1 or -9

**9th grade math**

I would divide by 4 to get m^2 - 6m + 9 = 0 , and now see it as a perfect square (m-3)^2 = 0 m-3 = 0 m = 3

**Calculus**

yes

**Calculus**

You are right to have those two asymptotes. look at http://www.wolframalpha.com/input/?i=plot+y+%3D+%28x%5E2%2B4x-5%29%2F%28x-2%29+%2C+y+%3D+x%2B6 What happens at the far right and the far left ?

**Math**

correct , there is only one choice, namely the 1

**College algebra**

white ones ---- x yellow ones ---- 31 - x solve for x 11.95x + 12.50(31-x) = 379.80

**precal**

I noticed that it factors quite easily using grouping f(x) = x^2(x - 2) - 15(x - 2) = (x-2)(x^2 - 15) so if (x-2)(x^2 - 15) = 0 then x-2 = 0 or x^2 - 15 = 0 x = 2 or x^2 - 15 = 0 x^2 = 15 x = ± √15 x^2 - 15 does not factor using rational numbers you could look it ...

**Math (Please Help Ms. Sue)**

The surface area of two similar solids is proportional to the square of their corresponding sides, so the square of the "doubling" is 2^2 or 4 So the surface area is quadrupled or let the original dimensions be l , w, and h surface area = 2lw + 2lh + 2 wh = 2(lw + lh...

**Calculus Help Please!!!**

a) very easy b) dy/dx = (π^x) * ln(π) c) tricky, first ln both sides lny y = ln(x^x) = xlnx y'/y = x(1/x) + lnx y' = x^x (1 + lnx) d) what is the derivative of a constant? e) xy = π x dy/dx + y = 0 dy/dx = -y/x

**Calculus Help Please!!!**

2x + 4x dy/dx + 4y + 2y dy/dx = 0 dy/dx(4x + 2y) = -2x - 4y dy/dx = -2(x+2y)/( 2(2x + y) = -(x+2y)/(2x+y) at (2,1) dy/dx = -4/5 so slope of tangent is 4/5 slope of normal is 5/4 tangent: y-1 = (-4/5)(x-2) y - 1 = (-4/5) + 8/5 y = (-4/5)x + 13/5 you do the normal is the same way.

**maths Pls help trig**

1. tanA = 1/3, and tanB = 1/7 I usually construct my triangles, (since the fraction gives you 2 of the sides of right-angled triangles, the third can always be found using Pythagoras) for tanA = 1/3, sinA = 1/√10 , cosA = 3/√10 for tanB = 1/7, sinB = 1/√50, c...

**MathS triG**

see solution above of the same question Please don't switch names.

**Algebra**

let her speed for the last two hours be x km/h total distance = 6(40) = 240 km 4(50) + 2(x) = 240 2x = 40 x = 20 speed of last part = 20 km/h

**college algebra**

So you are solving: 144 = 128t - 16t^2 16t^2 - 128t + 144 = 0 t^2 - 8t + 9 = 0 I will complete the square, (in this case faster than the formula) t^2 - 8t + 16 = -9+16 (t-4)^2 = 7 t-4 = ±√7 t = 4 ± √7 I get appr 6.65 and 1.35

**maths - HCF**

Did you notice that 119 = 7 x 17 , that is, its only factors are prime numbers. If we needed the HCF of 7 and 17 it would be 1 If we needed the LCM of 7 and 17 it would be 119 The product of 1 and 119 is 119 Our two numbers are 7 and 17 Further investigation: suppose we have t...

**math**

height/30 = sin 65 height = 30sin65° = .....

**Geometry**

A regular octagon forms 8 congruent isosceles triangles, where the central angles are 360/8° or 45° each. Your second part cannot be answered since you gave no data. Let the radius be r and consider one of the triangles.. If the base is 2s, then s/r = sin 67.5° s =...

**Math**

The way you worded it , we could form 3 letter words, or 5 letter words, etc. In other words, it becomes a "Scrabble" type question and not really a math question. Is that what you need ? or is it, "in how many ways can we arrange all the letters of the word ALG...

**Pre-Algebra**

I assume you meant 18 inches ( ' usually is the symbol for feet if I recall the imperial measure system) jenny --> x inches Demi--> half of that or x/2 inches Joel --> gets 3/4 of Jenny's or 3x/4 x + x/2 + 3x/4 = 18 times 4 4x + 2x + 3x = 72 9x = 72 x = 8 Jenn...

**Calculus Help**

let the height be h then the diameter of the base is 4h, thus the radius is 2h given: dV = 20 ft^3/min find: d(2h)/dt , when h = 12 V = (1/3)π r^2 h = (1/3) π (4h^2)(h) = (4/3) π h^3 dV/dt = 4π h^2 dh/dt 20 = 4π((144) dh/dt dh/dt = 5/(144π) so d(2...

**Math**

white / green = 3/2 w/18 = 3/2 2w = 54 w = 27 check: 27/18 = 3/2

**Math**

√(2x^2+5)=2 square both sides 2x^2 + 5 = 4 2x^2 = -1 x^2 = -1/2 no solution, since we cannot take the square root of a negative in the real number set.

**Algebra - Help!**

f(x)=(x^2+6x+5)/(x^2+3x−10) = (x+1)(x+5(/((x+5)(x-2) ) = (x+1)/(x-2) domain: any real value of x , x ≠ -5, 2 vertical asymptote: x = 2 hole: x = -5 horizontal asymptote: y = 1 see: http://www.wolframalpha.com/input/?i=plot+y+%3D+%28x%5E2%2B6x%2B5%29%2F%28x%5E2%2B3x...

**Algebra 2**

Looks like you label your a to be associated with the major axis (the axis containing the foci) In that case: b^2 + c^2 = a^2 b^2 + (√19)^2 = 7^2 b^2 + 19 = 49 b^2 = 30 so are correct about the centre, so (x+7)^2 /30 + y^2/49 = 1 http://www.mathwarehouse.com/ellipse/equa...

**uc-metc math16**

let the length of the chord be y let the horizontal distance of the kite be x ft given: dx/dt = 5 ?? (looks like a typo in the units) I will assume you meant: 5 mph = 5280 ft/60 min = 88 ft/min y^2 = x^2 + 40^2 when y = 50 50^2 = x^2 + 40^2 x = 30 2y dy/dt = 2x dx/dt dy/dt = 2...

**Calculus Help**

dy/dØ = cos(π/√(Ø^2+5) * (-1/2)(Ø^2 + 5)^(-3/2) ) * (2Ø) =(-πØ/(Ø^2 + 5)^(3/2) * cos(π/√(Ø^2+5) so when Ø = 2 dy/dØ = ... You do the button - pushing. Make sure to set your calculator to RAD

**Math**

1n + 7p = 5.55 3n + 2p = 12.28 from the first: n = 5.55-7p sub into the 2nd 3(5.55-7p) + 2p = 12.28 16.65 - 21p + 2p = 12.28 -19p = -4.37 p = .23 then n = 5.55 - 7(.23) = 3.94 check: 1 notebook + 7 pencils = 3.94 + 7(.23) = 5.55 3 notebooks + 2 pencils = 3(3.94) + 2(.23) = 12....

**Math**

Right now, (-2,-5) is 5 units below the x-axis so reflecting it over the x-axis would put it how many units above the x-axis ? And the total distance would be ...... ?

**foundation differential calculus**

I suspect a typo why would you have 2 terms in t^2 without adding them together? Is the last term supposed to be t^3 ? Clarify before I attempt it.

**Math**

sin 2x - sinx =0 2sinx cosx - sinx = 0 sinx(2cosx - 1) = 0 sinx = 0 or cosx = 1/2 for sinx = 0 x = 0, 180°, 360° for cosx = 1/2 x = 60° , 300° so for 0 ≤ x ≤ 360° x = 0, 60, 180, 300, 360 in radians: x = 0 , π/3 , π , 5π/3 , 2π

**Maths**

Did you make your sketch? Let the distance from the longer chord to the circle be x then the distance from the shorter chord to the circle is 17-x Let the radius be r then r^2 = x^2 + 12^2 and r^2 = (17-x)^2 + 5^2 so x^2 + 144 = (17-x)^2 + 25 x^2 + 144 = 289 - 34x + x^2 + 25 3...

**Math**

Ahh, looks like neither one of us read the question carefully. P(t) given the percentage that will FAIL, which is 30.2% But it says, "what percentage ... is usable " so 100 - 30.2 or 69.8%

**Math**

just plug in t = 3 P(3) = 100(1 - e^-.36) = 30.23 So appr 30.2%

**Math**

3(0.4h+5)>4(0.2h+7) 1.2h + 15 > .8h + 56 1.2h - .8h > 56 - 15 .4h > 41 h > 41/.4 h > 102.5

**math**

1 rotation = 360° = 2π radians at 7:00 the minute hand is on the 12 (straight up) and the hour hand has moved 7/12 of a rotation (7/12)(2π) = 7π/6 radians ---> or 210° so the smaller angle between the two hands = 2π - 7π/6 = 5π/6 radian...

**Math (check)**

Interpretation #1 : Slant height is the edge of the pyramid Consider the right-angled (internal) triangle formed by the height , half the diagonal of the base and hypotenuse being the slant height. let the slant height be x let the diagonal of the base be d d^2 = 1^2 + 12^2 = ...

**Algebra**

number of dach's --- x number of labs ----- 3x solve for x .... x + 3x = 200000

**Math**

I labelled the position of the church as C. angle C = 85° by the sine law: CP/sin35 = 600/sin85 CP = 600sin35°/sin85° = appr 345.5 m you do CQ

**Math**

Erin has 36 Julia has 2/3 as many as Erin ---> she has (2/3)(36) or 24 Julia also has 1/3 as many as Robert 24 = (1/3)(Robert) Robert has 72 Erin has 36 Julia has 24 Robert has 72 for a total of 132

**Algebra**

I suspect a typo and you meant: (x^2 - 25)/(x^2 - 3x - 10) , since that factors nicely into (x+5)(x-5)/( (x-5)(x+2) = (x+5)/(x-2) , x ≠ 5

**Math**

How about this pattern term(1) = 6 = 3(2^1) term(2) = 12 = 3(2^2) term(3) = 24 = 3(2^3) term(4) = 48 = 3(2^4) .... term(23) = .... = 3(2^23) term(n) = .....

**Math**

Q(t) = 200 (1/2)^(t/14.2) or Q(t) = 200(.5)^(.0704t) Q '(t) = (200/14.2) (.5)^(t/14.2) so when t = 21.5 Q ' (t) = (200/14.2) (.5)^(21.5/14.2) = 4.931 g/day

**algebra 2**

This produces an arithmetic series, where a = -5 and d = 8 sum(28) = (28/2)(-10 + 27(8)) = 2884

**Math**

I am sure you meant: Q(t) = 7000/(1 + 249 e^(-.6t) ) Q(10) = 7000/(1 + 249 e^(-.6(10) ) ) = appr 4328

**algebra 2**

so a = -12 , d = 5 sum(n) = (n/2)(2a + (n-1)d ) = (11/2)(-24 + 10(5)) = (11/2)(26) = 143 OR sum(n) = (n/2)(first + last) = (11/2)(-12 + 38) = 143

**algebra 2**

The famous mathematician Gauss, while he was in elementary school, notices the following 1+2+3+..+58+59+60 = (1+60) + (2+59) + (3+58) +.. (for a total of 30 of these pairs) = 30(61) = 1830

**MATH**

A polynomial must have a non-negative integer exponent. A and B contain square roots or exponents of 1/2 D: 11/(√x)^2 = 11/x or 11 x^-1 so none is a polynomial C: x - √11 yup, I would go with that one

**Algebra**

Did you mean: (2x+6)/(4x-8) ?? = 2(x+3)/(4(x-2)) = (x+3)/(2(x-2) or (x+3)/(2x - 4)

**Math**

I would consider it a "combination" , assuming that the order that you choose the appetizers should not matter Number of ways = C(4,2) x C(5,1) = 6x5 = 30

**Calculas**

P(4) = 100(4) - 3(16)ln(4) = .. I assume you have a calculator to get appr 333.46

**Calculus Help Please!!!**

dy/dx = 8x(1/2)(x-x^2)^(-1/2) (1 - 2x) + 8(x-x^2)^(1/2) = 4(x-x^2)^(-1/2) [ x(1-2x) + 2(x-x^2) ] = (4/√(x-x^2) ( x - 2x^2 + 2x - 2x^2) = (4/√(x-x^2) (3x - 4x^2) = 0 for a max/min 3x - 4x^2 = 0 x(3 - 4x) = 0 x = 0 ----> looks like it will yield a min x = 3/4 ---&...

**typo - Calculus Help Please!!!**

only valid for 0 ≤ x ≤ 1

**Calculus Help Please!!!**

only valid for 90 ≤ x ≤ 1 http://www.wolframalpha.com/input/?i=y+%3D+%3D+8x%28√%28x+−+x%5E2%29%29+%2C+0+%3C+x+%3C+1

**Algebra**

But you wanted to use the formula ... x = (-4 ± √100)/2 = (-4 ± 10)/2 = (-4+10)/2 or (-4-10)/2 = .....

**Algebra**

put it into standard form of a quadratic ... x^2 + 4x - 21 = 0 how nice that it factors .... (x+7)(x-3) = 0 so x = ... or x = ....

**percentages/word problem**

costprice = 80 after first markup, SP = 80(1.5) after 20% mark down = 8(1.5)(.8) another 15% markup = 8(1.5)(.8)(1.15) final markdown of 60% = 8(1.5)(.8)(1.15)(.4) = 8(.552) = 4.42

**Calculus1**

let the width of the beam be 2w and the depth be 2d S = k(2w)(4d^2) = 8k w(d^2) I drew a circle and incribed a rectangle with the defined dimensions. Notice the diagonal is 12 Construct a righ-angled triangle, with sides w, d, and hypotenuse 6 w^2 + d^2 = 36 d^2 = 36-w^2 in S...

**Math - Algebra2**

f(x)=3x+2 , g(x)=3/x^2+2 , h(x)=sqrt(x-2) (f∘g)(-5) to me means: f (g(x) ) so f(g(5)): g(-5) = 3/27 = 1/9 f(1/9) = 3(1/9) + 2 = 1/3 + 2 = 5/3 (f∘g)(-5) = 5/3 so the others the same way

**Precalculus**

w = -i - 2j describes the vector (-1,-2) so it the same way I showed you in the similar problem of unit length. YOu can do it in your head to get (-1/√5 , -2/√5)

**Algebra**

for any quadratic of the form y = ax^2 + bx + c the x of the vertex is -b/(2a) so for yours ... x of the vertex is -184/(-32) = 5.75 so it took 5.75 seconds, plug that into your original function to find the maximum height.

**Pre calculus**

correct notice you can check that the magnitude is 1 magnitude = √( (1/√2)^2 + (-1/√2)^2 ) =√( 1/2 + 1/2) = √1 = 1

**Pre calculus**

for any unit vector in the direction of (a,b) .... find the length of the vector = √(a^2+b^2) now divide each component by that magnitude to get (a/√(a^2+b^2) , b/√(a^2+b^2) or you could write it as (1/√(a^2+b^2) (a,b) apply this to your vector, let me ...

**algebra**

welcome

**algebra**

multiply each term by 2x(x+1) 6x - 2(x+1) = x(x+1) 6x - 2x - 2 = x^2 + x x^2 +3x + 2 = 0 (x-1)(x-3) = 0 x = 1 or x = 3

**Calculus1**

it would be a combination of the chain rule and the product rule, doing it as a single power the way I did it was easier as a product rule - chain rule for that term ..... x√x = x x^(1/2) derivative: x (1/2) x^(-1/2) + (1)x^(1/2) = x/(2√x) + √x = x/(2√x...

**Calculus1**

yes, of course, silly me, the exponent on the last one was 3/2

**Calculus1**

I would rewrite as h(x) = 3x^(1/2) - x^(3/2) h ' (x) = (3/2) x^(-1/2) - (2/3) x^(1/2) = 3/(2√x) - (2/3) √x

**Algebra**

it will hit the ground when the height h(t) = 0 -16t^2 + 144 = 0 16t^2 = 144 t^2 = 144/16 = 9 so t =√9 = 3 after 3 seconds

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