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Thank you this helped me get the correct answer when plugging in values for \(k_1, k_2\) and \(v_f\)

I still struggle to understand how did you solve to find these x(t) and v(t)?

For each sequence, decide whether it could be arithmetic, geometric, or neither.

\(\begin{array}{|l|l|c|} \hline 1. & 25,5,1,\ldots & \text{geometric} \quad r=\dfrac{1}{5} \\ \hline 2. & 25,19,13,\ldots & \text{arithmetic} \quad d=-6 \\ \hline 3. & 4,9,16,\ldots \\ & 4,10,16,\ldots & \text{arithmetic} \quad d=6 \\ & 4,8,16,\ldots & \text{geometric} \quad r=2 \\ \hline 4. & 50,60,70,\ldots & \text{arithmetic} \quad d=10 \\ \hline 5. & \dfrac{1}{2},3,18,\ldots & \text{geometric} \quad r = 6 \\ \hline \end{array}\)

I think the solutions can be written as:

\(v=\sqrt{\frac{k_1}{k_2}}\tanh{(\sqrt{k_1k_2}t)}\text{ and }x=\ln{(\cosh{(\sqrt{k_1k_2}t)})}/k_2\)

which I obtained by guessing a few forms for the velocity and manipulating them until they matched numerical solutions for all the values of k1 and k2 that I tried!!   I've no doubt there is a more straightforward way of doing it!!

\sqrt[3]{24a^4b^6c^{11}}

\(\sqrt[3]{24a^4b^6c^{11}}\\ =\sqrt[3]{8*2*a^3*a*b^3*b^3*c^{9}*c^2}\\ =\sqrt[3]{2^3*a^3*b^3*b^3*c^{9}*2ac^2}\\ =2*a*b*b*c^3*\sqrt[3]{2ac^2}\\ =2ab^2c^3\sqrt[3]{2ac^2}\\ =2^1a^1b^2c^3\sqrt[3]{2ac^2}\\\)

The way I interprete the question the sum of the exponents of the variables that are outside the radical is 7

I expect you forgot the ones.    

Hi CalculatorUser,

A big thanks to you and Rom.

To answer your question... Yes it does.

y= x^2+8x     is a concave up parabola.   (so it will have a minimum value)

y=x(x+8)

So the 2 roots are at x=0 and  x=-8

The axis of symmetry is half way between them  x=(0+-8)/2 = -4

The minimum will be the y value when x=-4

y=-4(-4+8) = -4*4 = -16

sketch what I have done so you can see what I am talking about. 

You are right. Not everyone uses this site to cheat from.

I am sure some people learn a great deal here.

If I did not believe that then I would not be here. 

Next time make sure you delete all the meaningless $ signs.  So that your question presents properly.

\(\text{Let F(x) and G(x) be polynomials such that }\\ f(x)=F(x)(x+3)(x-4)(x-8)\\ g(x)=G(x)(x+3)(x-4)(x-8)(x+5)(x-2)\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+3)(x-4)(x-8)(x+5)(x-2)}{F(x)(x+3)(x-4)(x-8)}\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+5)(x-2)}{F(x)}\\ \)

g(x) is only divisable by f(x) if F(x) is a factor of G(x)(x+5)(x-2)

for instance in the following example g(x) is not divisable by f(x)

\(g(x)=7(x+5)(x-2)(x+3)(x-4)(x-8)\\ f(x)=(x^8+x^5-4x)(x+3)(x-4)(x-8)\\\)

Express the following in simplest form : \((-2)^{4x+2}(2)^{6x-5}(8)^x\)

\(\begin{array}{|rcll|} \hline && \mathbf{(-2)^{4x+2}(2)^{6x-5}(8)^x} \\ &=& \Big((-1)(2) \Big)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{(ab)^c = a^cb^c} \\ &=& \left(-1 \right)^{4x+2}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{a^{b+c} = a^ba^c} \\ &=& \left(-1 \right)^{4x}\left(-1 \right)^{2}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{ (-1)^2 = (-1)(-1)=1 } \\ &=& \left(-1 \right)^{4x}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad 8 = 2^3 \\ &=& \left(-1 \right)^{4x}\left( 2 \right)^{4x+2}(2)^{6x-5} \left(2^3 \right)^x \quad & | \quad \boxed{ \left(a^b \right)^c = a^{bc} \\ \left(2^3 \right)^x = 2^{3x} } \\ &=& \left(-1 \right)^{4x}\left( 2 \right)^{4x+2}(2)^{6x-5} (2^{3x}) \quad & | \quad \boxed{ a^ba^ca^d=a^{b+c+d} } \\ &=& \left(-1 \right)^{4x}2^{4x+2+6x-5+3x} \\ &=& \left(-1 \right)^{4x}2^{4x+2+6x-5+3x} \\ &\mathbf{=}& \mathbf{\left(-1 \right)^{4x}2^{13x-3}} \\ \hline \end{array} \)

If $a\equiv 18\pmod{42}$ and $b\equiv 73\pmod{42}$, then for what integer $n$ in the set $\{100,101,102,\ldots,140,141\}$ is it true that $$a-b\equiv n\pmod{42}~?$$

Help please!

\(\begin{array}{|rcll|} \hline a &\equiv& 18\pmod{42} \\ b &\equiv& 73\pmod{42} \\ \hline a-b &\equiv& 18- 73 \pmod{42} \\ a-b &\equiv& -55 \pmod{42} \\ a-b &\equiv& -55+42+42 \pmod{42} \\ a-b &\equiv& 29 \pmod{42} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a-b &\equiv& 29 \pmod{42} \\ a-b &=& 29 + z\cdot 42,\ z \in \mathbb{Z} \\ && \begin{array}{|r|c|c|} \hline & \mathbf{n} = (a-b)\pmod{42} & \\ z & =29 + z\cdot 42 & n \text{ in set} \\ \hline 0 & n=29 & \\ \hline 1 & n=71 & \\ \hline 2 & n=113 & \checkmark \\ \hline 3 & n=155 & \\ \hline \ldots \\ \hline \end{array} \\ \hline \end{array}\)

It is true for \(n = 113 \)

Find the remainder when \(3 \times 13 \times 23 \times 33 \times \ldots \times 183 \times 193\) is divided by \(5\).

\(\begin{array}{|rcll|} \hline 3 \pmod{5} &\equiv& 3 \\ 13 \pmod{5} &\equiv& 3 \\ 23 \pmod{5} &\equiv& 3 \\ 33 \pmod{5} &\equiv& 3 \\ \ldots \\ 183 \pmod{5} &\equiv& 3 \\ 193 \pmod{5} &\equiv& 3 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{3 \times 13 \times 23 \times 33 \times \ldots \times 183 \times 193 \pmod{5}} \\ &\equiv& 3 \times 3 \times 3 \times 3 \times \ldots \times 3 \times 3 \pmod{5} \\ &\equiv& 3^{20} \pmod{5} \quad | \quad 3^2\pmod{5} \equiv 9 \pmod{5} \equiv -1 \pmod{5} \\ &\equiv& \left(3^2\right)^{10} \pmod{5} \\ &\equiv& \left(-1\right)^{10} \pmod{5} \\ & \mathbf{\equiv}& \mathbf{1 \pmod{5}} \\ \hline \end{array}\)

The remainder is 1

Let the acreage reseved for corn =C
The acreage reserved for wheat   =2C
The acreage reserved for soybeans=1,300 - C - 2C
45C + 120C + 50[1,300 - C - 2C] = $69, 200, solve for C
C = 280 - acres reserved for corn
280 x 2 =560 acres reserved for wheat
1300 - 280 - 560 =460 acres reserved for soybeans.

your answer was correct, and guessing has it's place in solving problems,

but there was a concrete way to solve this one.

distance/rate = time

1/5 km  /  1/12 km/hr  =  12/5 hr  = 2 hr 24 min

5+6rt2

Roots   are     5+- sqrt(25+4(a)(3)     /    2a   

[5 + sqrt(25+12a) - 5 + sqrt(25+12a) ] /2a  = sqrt61/3

       2 sqrt (25+12a)/2a = sqrt(61)/3

                  sqrt(25+12a)/a = sqrt(61)/3    results in a = 3

hint    :      3^x   *    3^4   =  3^(x+4)   not   3^(4x)

This is an algebra problem that can be solved by making an equation with 3 variables.

(1) Make Equations

First equation: "It costs $45 per acre to grow corn, $60 to grow wheat and $50 to grow soybeans" = "A farmer has 1,300 acres of land on which he grows corn, wheat, and soybeans."

So the first equation is this  : \(x+y+z=1300\) the three variables representing HOW MANY acres

Second equation : "It costs $45 per acre to grow corn, $60 to grow wheat and $50 to grow soybeans" = "He has allocated $69,200 for the cost of growing his crops"

So the second equation is this : \(45x+60y+50z=69200\)

Third equation : "Because of market demand the farmer will grow twice as many acres of wheat as of corn."

So the third equation is this : \(y=2x\)

(2) Solve

Now we know the two equations, we can solve.

We can substitute the third equation into the first equation to get \(3x+z=1300\)

Now isolate z.

\(z=1300-3x\)

We can also substitute the third equation into the second equation.

\(165x+50z=69200\)

Now substitute the equation \(z=1300-3x\) into the equation above.

After all this, you should get the answer of 

\(\boxed{x=280}\)

\(\boxed{y=560}\)

\(\boxed{z=460}\)

So in the end, we should have 280 acres of corn, 560 acres of wheat, and 460 acres of soybeans.

8x6 = 48 6x8=48

Yes.  This is called the commutative property of multiplication.   

a x b = b x a

I think part of your awsnere did not render.  I just have blank spaces

Sorry, same for this one.

Sorry, cannot read your "LaTex".

product(3, 13, 23 , 33 , 43 , 53 , 63 , 73 , 83 , 93 , 103 , 113 , 123 , 133 , 143 , 153 , 163 , 173 , 183 , 193)  mod 5

 = 1  

Since if the turtle reaches the pond the distance away is 0,  we can make the equation \(\frac{1}{5}-\frac{1}{12}x=0\)

Solving we get 2.4 hours, or 144 minutes