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1/24  +  1/36 ==5/72
Bobby and Danny working together can finish the job in: 72 / 5 ==14.4 days.

8 / 14.4 =5/9 of the work has already been finished.

1 - 5/9 ==4/9 of the work remains to be finished by Andy. Since Andy can finish the job, working alone, in 27 days, therefore he can finish:

4 / 9 (of the remaining work) x 27 days ==108 / 9 ==12 days.

n = 21.

Yes, they are the same BUT they have to be included.

I made a mistake in my original answer.

The point (5,4) is NOT a point on the graph  y=floor of x   because the floor of 5 is 5 (not 4)

So there are only 4 points of intersection

Here is the correct graph

Thank you Melody! According to your information, I covered

f(x)= 0.8x with g(x)=ceil(x-1) in my graph program. I hadn't used ceil before.

There are three points of intersection and two points of contact.

Question: Do points of intersection and points of contact have the same designation in the English-speaking world?

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Sorry asinus, I didn't see your question earlier.

[x] means nothing to me either.

I looked it up and it seems to be used sometimes in place of the floor function. \(\lfloor x\rfloor\)

I assume that this is because the floor function symbol is not on an ordinary keyboard and it is not a common symbol.

The floor function simply means 'round down' to the nearest integer.

so for example:

\(\lfloor 3 \rfloor =3\\ \lfloor 3.2 \rfloor =3\\ \lfloor 3.7 \rfloor =3\\ \lfloor 3.99999 \rfloor =3\\ \lfloor 4 \rfloor =4\\\)

 What is the measure of angle DCA ?

Hello Guest!

Im Dreieck ABC ist der Winkel ABC = 2 \alpha und der Winkel ACB = 148 - \alpha. D ist ein Punkt auf der Strecke AB, so dass CB = BD . Was ist das Maß (in Grad) des Winkels DCA?

\(\alpha +2\alpha+148°-\alpha =180°\\ \alpha =16°\)

\(∠DCA=148°-\alpha-\dfrac{180°-2\alpha}{2}=148°-16°-\dfrac{180°-2\cdot 16°}{2}\)

\(∠DCA=58°\)

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What is AC?

Hello Guest!

Point A is the origin of an x-y coordinate system.

\(f(x)=-tan(\ 60^\circ)\cdot x\\ x_D=-46\cdot cos(\ 60^\circ)=-23\\ y_D=46\cdot sin(\ 60^\circ)=39.837\)

\(g(x)=tan(\ 30^\circ)\cdot (x+23)+39.837\\ x_C=13\\ g(13)=y_C=tan(\ 30^\circ)\cdot (13+23)+39.837\\ y_C=60.622\)

\(\overline{AC}=\sqrt{13^2+60.622^2}\)

\(\overline{AC}=62\)

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What point must the graph of g pass through? Function f passes through (7,-1).

Hello Guest!

f(x) = x - 8

    x      |  -4    -3    -2    -1    0   1   2   3   4 

f(x) = z | -12 -11  -10   -9   -8  -7  -6  -5  -4

g(z) = z + 8

g(f(x))  |  -4    -3    -2    -1    0   1   2   3   4

Point (-7,1) must the graph of g pass through.

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I thought that was abs(x). Please explain to us what the graph means. I didn't know anything like that before.

Assuming that   [x] is a reference to the floor function,    which is normally written as  \(\lfloor x \rfloor\)

then there are 5

There is only one touch point (0,0), no intersection point.

  !

Thanks for that great interaction.

I think you have drawn the triangle correctly.

It does not work out to be isosceles.

I let the two unknown lengths be x and y

then I formed two equations using the cosint rule, one beginning with  26^2=    and the other with  7^2=

then I solved them simultateously (with subtraction) to get an expression for y in terms of x.

then i subed back into an original equation to get two numbers for x, 

[One is actually x and the other is y becasue the terms are interchangeable] 

See how much that helps.

* Note that my method may not be the best method.

Solving, \(x = -1\) and \(y = 2\). 

Thus, \(-1 - 2 = \color{brown}\boxed{-3}\)

1 - Compounded annually:  $2 ==$1  x  1.12^n, solve for n, where n == number of years.

2 ==1.12^n, take the log of both sides

n ==log(2) / log(1.12)

n==6.12 years.

2 - Compounded monthly:  $2 ==$1  x  [ 1  +  0.12/12]^n, where n==number of months.

2 ==1.01^n, take the log of both sides

n ==log(2) / log(1.01)

n ==69.66 months / 12 ==5.81 years.

3 - Compounded daily:   $2==$1  x  [1  +  0.12/365]^n, where n==number of days

2 ==1  x  1.00032876712^n, take the log of both sides.

n ==log(2) / log(1.00032876712)

n ==2,108.67 days / 365 ==5.78 years.

4 - Compounded continuously:  $2 == $1  x  e^(0.12n)

2 ==e^(0.12n), take the log of both sides

0.12n ==ln(2) ==0.69314718....

n ==0.69314718.... / 0.12 ==5.7762265 ==~5.78 years.

x - y = -5.

Hello! Thank you for the tip!! So far. I have drawn a large triangle with a similar smaller triangle inside. I will attemp to upload the picture, however I am new to the forum and am not sure of the tools used. for reference, AED is the larger triangle will BEC is the  smaller and similar triangle inside. I assumed that it was isoceles, but i am probably wrong 🥲. Because I am at an ameuteur geometry level, I attempted to solve this problem with similar triangles, but that led to the solution of 77/19, which I am decently sure is incorrect. Any help is appreciated. Thanks again!

\(33 \frac{1}{3} \% = 33.3\overline{3} \% = \frac{1}{3}\)

\(\frac{1}{3} \cdot 12 = \frac{12}{3} = 4\)

The first thing you have to do is draw a pic.

You probably won't get it right the first time.

You have to tweak it, and redraw it, a few times until it is right.

Have you done that?

If so, upload a pic of  it, or just describe what you pic looks like.

after that you can use cosine rule twice and then solve them the best you can simultaneously

I won't say more until you have contributed something of your own.

What point must the graph of g pass through? Function f passes through (7,-1).

Hello Guest!

f(x) = x - 8

    x      |  -4    -3    -2    -1    0   1   2   3   4 

f(x) = z | -12 -11  -10   -9   -8  -7  -6  -5  -4

g(z) = z + 8

g(f(x))  |  -4    -3    -2    -1    0   1   2   3   4

Point (-7,1) must the graph of g pass through.

  !

There is not enough information given to find and area.

Below are 2 of and infinite number of quadrilaterals that meet this description.

It is visually clear that the one on the right has a bigger area than the one on the left

try dividing

minus 20 from both sides

Anyway, if you are interested enough to respond I will discuss my thoughts with you.