paulmcquad

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4 years, 62 days

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These are questions asked by paulmcquad

My question is can the last step be equal to 1/6.

I want to output to 1/6. it outputs to sqrt(9)/18 - (See Below).

NULL"5)Square root: undefined - DNE (Limit does not exist):"

NULL

NULL

NULL

limit((sqrt(x+1)-3)/(x-8), x = 8)"(=)"1/6 

 

 

limit((sqrt(x+1)-3)/(x-8), x = 8)Limit(((x+1)^(1/2)-3)/(x-8), x = 8) = (1/18)*9^(1/2)"(=)"Limit(((x+1)^(1/2)-3)/(x-8), x = 8) = 1/6

NULL

Download limit-sqrt-5.mw

I'm trying to graph this function but it comes out a straight line.

Thanks in Advance.

plot(cos(Pi/4), -2*Pi .. 2*Pi)

plot(cos((1/4)*Pi), -2*Pi .. 2*Pi)

 

``


 

Download Graph-cos.mw

Hi i'm trying to plot a height(feet) and distance(time) of a projectile.

I doesn't seem to land.

Thanks in Advance

 

Projectile.mw

Hi,

I'm trying to solve for Double Inequality. (1.1) and (1.3) combined.

Thanks in Advance
 

Example: Solving a Double Inequality

 

-3 <= 6*x-1

0 <= 6*x+2

(1.1)

"(->)"

[[-1/3 <= x]]

(1.2)

6*x-1 < 3

6*x < 4

(1.3)

"(->)"[[x < 2/3]]

"ex1(x):=6 x-1"

proc (x) options operator, arrow, function_assign; 6*x-1 end proc

(1.4)

"->"

 

"->"

"ex2(y):=3"

proc (y) options operator, arrow, function_assign; 3 end proc

(1.5)

"ex3(y):=-3"

proc (y) options operator, arrow, function_assign; -3 end proc

(1.6)

NULL


 

Download Double_Inequality.mw

Hi,

I was wondering is there an easier way for me to solve an Equation Involving an Absolute Value?


 

Example: Solving an Equation Involving an Absolute Value

 

x^2-3*x = -4*x+6

x^2-3*x = -4*x+6

(1.1)

"(->)"

[[x = 2], [x = -3]]

(1.2)

-x^2+3*x = -4*x+6

-x^2+3*x = -4*x+6

(1.3)

"(->)"

[[x = 1], [x = 6]]

(1.4)

"ex7(x):=|x^(2)-3 x|+4 x-6"

proc (x) options operator, arrow, function_assign; abs(x^2-3*x)+4*x-6 end proc

(1.5)

"->"

 

``

NULL


 

Download Absolute_Value.mw

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