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How many adults got off the ferry?

Hello Guest!

\(f=a+g+b\\ a=0.75f\\ 0.25f=b+g\\ a=b+168\\ g=0.4\cdot 0.25f\\ g+b=0,4(f-168)\)

\(g+b=0.25f=0.4(f-168)\\ 0.25f=0.4f-67.2\\ 0.15f=67.2\\ f=448\\ \color{blue}a=336\)

336 adults got off the ferry.

No!

I have to think again.

  !

A sequence is defined by 

\(\begin{array}{|rcll|} \hline t_{345} &=& 2 &\quad 345 = 4*86+1 \\ \hline \end{array}\)

\((6x + 1/(3x) + 2)^6=46656x^6+\dfrac{1}{729x^6}+\ ...\ +9792x+\dfrac{544}{x}+\color{blue}2624\)

  !

 Add 6 to both sides:                          \(2x^2+8x+6\le 0\)

Divide by 2:                                         \(x^2+4x+3\le0\)

Factorise:                                            \((x+3)(x+1)\le 0\)

Critical values:                                     \(x_1=-3,\text{and, }x_2=-1\)

Sketch the Quadratic, it is open-upwards and intersects the x-axis at x=-3 and x=-1.

Thus, the desired region is below the x-axis that is, between -3 and -1

Therefore,                                \(-3\le x\le-1\) is the desired solution.

In interval notation:   \(x \in[-3,-1]\)

Find the general term of the following sequences:

a)\( 7, 13, 19, 25, 31, \ldots\)
\(a_n=6n+1\)

b) \(4, –6, 9, -\frac{27}{2}, \ldots\)
\(a_n=4*(-\frac{3}{2})^{n-1} \\ a_n=a_{n-1}*(-\frac{3}{2})\)

Let the amount of money she started with==M

M - 18 - what she had left after buying some English books.

5/12[M - 18] - what she had left after buying  some Chinese books

5/12[M - 18] ==1/3M, solve for M

M ==$90 - amount of money she started with

7/12[90 - 18] ==$42 - she spent buying Chinese books.

split the bath into 5 parts.

2/5 will be cold water and 3/5 will be hot water.

Hot  23*3/5 = 13.8 minutes       

Cold 17*2/5 = 6.8 minutes

13.8-6.8 =7 minutes

Let \(r = \) the radius of the cylinder, and let \(h =\) the hieght of the cylinder.

The formula for the volume of a cylinder is \(\pi r^2 h\).

We are given the diameter, which is \(2 \times r\). This means that  \(r = 16\)

Now, we can subsitute what we know, to find the volume: \(3.4 \times 16 \times 15 \)

Can you take it from here?

We have: \(\large{1 \over {1 + {1 \over {\sqrt3 + 2}}}}\)

Lets start with \(\large{1 \over {\sqrt3 + 2}}\). To rationalize the denominator, multiply the fraction by \(\large{{ \sqrt 3 - 2} \over { \sqrt 3 - 2}}\). By doing this, we keep the value of the fraction the same, while keeping square roots out of the denominator, because we are, effectivelty, multiplying by 1. The denominator of the fraction becomes \({(\sqrt 3 + 2) ( \sqrt 3 - 2) = 3 - 4 = -1}\). However, the numerator is still \(\large{\sqrt 3 -2} \), because anything multiplied by 1 is itself. 

Now, the fraction is: \(\large {{ \sqrt 3 - 2 } \over -1}\). Multiplying by \(\large{ -1 \over -1}\), we get: \(2 - \sqrt 3 \) remember, we aren't changing the value of the fraction by doing this, because \({ -1 \over -1} =1\).

Adding 1 to this gives us \(3 - \sqrt 3 \). 

This means we have the fraction \(\large{1\over { 3 - \sqrt 3}}\). 

Now, we have to rationalize this again. 

Can you take it from here?

Recall the quadratic formula: \(\large{x = {-b \pm \sqrt{b^2-4ac} \over 2a}}\).

Subsituting in what we know, we get: \(\large{{-99 \pm \sqrt{8021}} = {-99 \pm \sqrt{99^2-4\times{1 \over 2}c} \over 2\times {1 \over 2}}}\)

Notice the discriminant (\(b^2 - 4ac\)). It must equal 8021, because the denominator is 1. 

This means that we can derive the equation: \(99^2 - 2c = 8021\). 

Now, we have to solve for c. 

Can you take it from here?

The equation of the circle is \(x^2 + k^2 = 1\). 

We know that \(x = {1 \over 2}\). Subsituting this in gives us: \({1 \over 2} ^2 + k^2 = 1\)

This means \(k^2 = {3 \over 4}\).

Now, we have to take the square root, and round the result to the nearest hundredth. 

Note that \(m\) is essentially the largest prime number less than 100. To find this, count down odd numbers from 100, and count how many factors they have.

Also note that \(n\) must be a perfect square, because that is the only way for a number to have 3 factors. 

Can you take it from here?

I don't think there is a trick as such.  It is just getting your to think about numbers, which is what you have done,

thanks for the answer. I have a test coming up and this helped a lot

8643*975

9643*875

8753*964

9753*864

what is the trick to the problem?

LCM{1!+2!+3!,  2!+3!+4!,  3!+4!+5!} ==LCM{9,  32,  150} ==7,200

7,200 ==10 * 6!

10 + 6 ==16

Although 661, 667, 669 are possibilities with regard to the 2-six requirement, they can't be 

part of the solution because the problem says "How many even three-digit positive integers"  

_.

At least how many vowels?

Substitute h(y) = 5, and see what the value of y is. That value is h^(-1) (5).

\(5 = \dfrac{1 + y}{2 - y}\\ 10 - 5y = 1 + y\\ 6y = 9\\ y = \dfrac32\)

Therefore \(h^{-1}(5) = \dfrac32\).

Just substitute a pair of (p, s) into the equations in the options and see which one is always true.

For example, take the pair (p, s) = (13, 1) and the option \(p = \dfrac s9 + 4\). When we substitute it into the equation, it is not right, since 13 is not 1/9 + 4. So this option is not the correct answer. Repeat this until you find the one that will give you a correct equation.

I think you forgot to close your brackets.

Simplifying from the innermost brackets:

\(y + 2\left(y + 3\left(y + 4 \left(y + \dfrac1{24}\right)\right)\right) = 100\\ y + 2 \left(y + 3 \left(5y + \dfrac16\right)\right) = 100\\ y + 2 \left(16y + \dfrac12\right) = 100\\ 33y +1 = 100\\ y = 3\)

I'm doing fine, thank you Guest.

Just copy this problem and google it, and you will have the answer.

Google result: https://web2.0calc.com/questions/plz-help_2147

Note that\(g(x) = \dfrac2{2 + 4x^2} + \dfrac8{1 + x^2} = \dfrac1{1 + 2x^2} + \dfrac8{1 + x^2}\).

Also, \(x^2 + 1 \geq 1\) and \(1 +2x^2 \geq 1\).

Then, \(0 < \dfrac1{1 + 2x^2} \leq 1\) and \(0 < \dfrac1{1 + x^2} \leq 1\).

That means \(0 < \dfrac8{1 + x^2} \leq 8\).

Therefore, \(0 < g(x) \leq 9\) by adding the inequalities. Can you rewrite the range as an interval?