if anyone can help with any of these problems, that would be great :)

[math4lyfe]

1. solve: x+3/|x-2|>0

2. The sides of a parallelogram have lengths 18 and 26 ft. and one angle of 39 degrees. Find the length of the longer diagonal.

4. Find a cubic function f(x)=ax³+bx²+cx+d that has a local maximum value of 3 at -2 and a local minimum value of 0 at 1.

5. If y=asin(cx) + bcos (cx), where a, b, and c are constants, then d²y/(dx²) is...




PLEASE HELP!

 

[tkhunny]

I REALLY hope your exam isn't soon, because if you REALLY can't do ANY of these, you're toast.

1) Please remember your Order of Operations. x+3/|x-2| is NOT the same as (x+3)/|x-2|

Notice how the latter expression does not exist for x = 2.

After that, observe that the absolute values in the denominator are, by defintion, positive.

The only consideration for the sign of the expression is (x+3). Where is that positive?

 

[anandcu3]

5) dy/dx = ac Cos(cx) - bcsin(cx)
d^2y / dx^2 = -ac^2sin (cx) - ac^2Cos (cx)
= -c^2 (aSin(cx) + b Sin(cx) )
Ans =-c^2 y

 

[galactus]

4. Find a cubic function f(x)=ax³+bx²+cx+d that has a local maximum value of 3 at -2 and a local minimum value of 0 at 1.

You have four variables to solve for: a,b,c,d.

Thus, you're going to need four equations.

These can be found from the given info.

f(x) is equal to 3 at x=-2.

Plugging in x=-2, we have \(\displaystyle -8a+4b-2c+d=3\)

It is equal to 0 at x=1:

\(\displaystyle a+b+c+d=0\)

Now, for the derivatives.

There is a max at x=-2. This means the derivative is 0 there.

\(\displaystyle f'(x)=3ax^{2}+2bx+c=0\)

Plug in x=-2:

\(\displaystyle 12a-4b+c=0\)

Also, at x=1:

\(\displaystyle 3a+2b+c=0\)

There are four equations with four unknowns:

\(\displaystyle -8a+4b-2c+d=3\)

\(\displaystyle a+b+c+d=0\)

\(\displaystyle 12a-4b+c=0\)

\(\displaystyle 3a+2b+c=0\)

Solve for a,b,c, and d.

 

[soroban]

Hello, math4lyfe!

2. The sides of a parallelogram have lengths 18 and 26 ft. and one angle of 39 degrees.
Find the length of the longer diagonal.

          S o  *  *  *  *  *  *  o R
           *                    *
          *                    *
         *                    * 18
        *                    *
       * 39d           141d *
    P o  *  *  *  *  *  *  o Q
               26

We have parallelogrm \(\displaystyle PQRS\!:\;PQ = 26,\;QR = 18.\)

. . \(\displaystyle \angle P = 39^o \quad\Rightarrow\quad \angle Q = 141^o\)


Draw diagonal \(\displaystyle PR\).

Apply the Law of Cosines to \(\displaystyle \Delta PQR:\)

. . \(\displaystyle PR^2 \:=\:26^2 + 18^2 - 2(26)(18)\cos141^o\)

 

[math4lyfe]

You have four variables to solve for: a,b,c,d.

Thus, you're going to need four equations.

These can be found from the given info.

f(x) is equal to 3 at x=-2.

Plugging in x=-2, we have \(\displaystyle -8a+4b-2c+d=3\)

It is equal to 0 at x=1:

\(\displaystyle a+b+c+d=0\)

Now, for the derivatives.

There is a max at x=-2. This means the derivative is 0 there.

\(\displaystyle f'(x)=3ax^{2}+2bx+c=0\)

Plug in x=-2:

\(\displaystyle 12a-4b+c=0\)

Also, at x=1:

\(\displaystyle 3a+2b+c=0\)

There are four equations with four unknowns:

\(\displaystyle -8a+4b-2c+d=3\)

\(\displaystyle a+b+c+d=0\)

\(\displaystyle 12a-4b+c=0\)

\(\displaystyle 3a+2b+c=0\)

Solve for a,b,c, and d.

you are going to think i'm so dumb...but how do you solve for four different equations?

 

[galactus]

Are you allowed to use a calculator?. Do you have one?.

If so, many solve systems of equations.

You can also do it by hand by using the elimination method.

i.e you could multiply the second row by -1 and add to the first row. This eliminates d and you have:

\(\displaystyle \begin{array}-9a+3b-3c=3\\ 12a-4b+c=0\\ 3a+2b+c=0\end{array}\)

Have you seen the elimination method before?. You certainly should have before you got to calculus.

 

[math4lyfe]

Are you allowed to use a calculator?. Do you have one?.

If so, many solve systems of equations.

You can also do it by hand by using the elimination method.

i.e you could multiply the second row by -1 and add to the first row. This eliminates d and you have:

\(\displaystyle \begin{array}-9a+3b-3c=3\\ 12a-4b+c=0\\ 3a+2b+c=0\end{array}\)

Have you seen the elimination method before?. You certainly should have before you got to calculus.

Yes, of course I've seen the elimination method haha. I just wasn't sure how to do it with four equations, So you find a b c and d and plug it back in to the original equation?