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Where are you getting these word problems

Let J be the number of books that Jonson has and let M be the number of books mandy has.

J-5=M+5

3(M-4)=(J+4)

now solve them simultaneously.  Can you do that?

What do you get for M and J?

if so, then it would be:

\(6\cdot 3\cdot [-9+4(5\cdot 3-9)]-3\cdot (40-8)\)

(calculate parenthesis) \(\left[-9+4\left(5\cdot3-9\right)\right]=15\)

\(6\cdot3\cdot15-3\left(40-8\right)\)

(calculate parenthesis) \((40-2)=38\)

\(6\cdot3\cdot15-3\cdot32 \)

(multiplication) \(6\cdot3\cdot15=270\)

\(270-3\cdot32\)

(multiplication) \(3\cdot32=96\)

\(270-96=174\)

are the dots multiplication??

3, 5, 9 is not a valid triangle since 3 + 5 < 9

=^._.^=

(2/3)/2 = 1/3

She only needs 1/3 cups of flour. 

=^._.^=

The table isn't showing up, but the theoretical probability of spinning a 5 is 1/10*100 = 10.  

=^._.^=

I'm sorry, but is this a problem on the map test?

=^._.^=

Oh thanks

Find the vertex of the graph of the equation y=-2x^2+18x-15.

Hello Guest!

\(f(x) = ax² + bx + c\\ y=-2x^2+18x-15\)

\(x_v=-\frac{p}{2}=-\frac{\frac{18}{-2}}{2}\\ \color{blue}x_v=4.5\)

\(y_v=q-\frac{p^2}{4}=\frac{-15}{-2}-\frac{(\frac{18}{-2})^2}{4}\\ y_v=-2x_v^2+18x_v-15=-2\cdot 4.5^2+18\cdot 4.5-15\\ \color{blue}y_v=25.5\)

Informed at:

Thanks very much Heureka.     

Hopefully I will remember next time.  ://

Continued in response to this post:

The sample has 25 fish, 5 from each tank. The ration of fish rot in the sample is \(\frac{4}{25}.\) The store has a total of 5(40) = 200 fish, the population the sample of 25 comes from. The assumption is that if the sampling is done randomly the ratio of fish rot in the population would be the same as that of the sample. So we can expect\(\frac{4}{25}(200)=32\) fish with fish rot in the whole store.

see the previous problem; The answer should be \(y={(x-1)}^{2}-4,\ in\ the\ vertex\ form,\ and\)\(y={x}^{2}-2x-3,\ in\ the\ standard\ form.\)

There is a falsity in the statement of this problem; a is definitely not an integer. It is even evident from the graph that a is around \(\frac{2}{3}\), since if you start with the vertex and move one unit to the right the y value increases by less than a unit; b and c are, however integers, b = -4 and c =7. You can find these values by writing the vertex form \(y=a{(x-h)}^{2}+k\) and plugging in 3 for h and 1 for k and expanding to get \(y = a({x}^{2}-6x+9) +1\). Finally, since the point (0, 7) is on the graph, we have 7 = 9a +1 , which yields \(a =\frac{2}{3}\). So \(b=\frac{2}{3}(-6)=-4\) and \(c=\frac{2}{3}(9) + 1= 7\). Finally, the parabola has equation

\(y=\frac{2}{3}{x}^{2}-4x+7.\)

Ok, I have checked your answer and it works so now I have to go back and find my error.

Refer back to the original question.  

I've made a correction post.

This is the original post:

Nice work guest  :)

If G is the midpoint of DC, the the three triangles BCG, BGD, and BDA have the same area, having the same height BH and equal bases. The two triangles ADE and AEB also have the same area since they have the same height AH' and equal bases. So it follows that the area of AEB is \(\frac{1}{6}\) the area of the triangle ABC, or \(\frac{200}{3}\) square centimeters.

Hi Yeager,

It is nice to see you trying to make a positive contribution.

the interior angles of a square add up to  90*4 = 360

the exterior angles add up to   270*4 = 1080 degrees 

How many sides has an equiangular polygon if the sum of its exterior angles is equal to the sum of its interior angles?​

MMm

I think that is impossible... 

The interior angles have to be less than 180 degrees and the exterior angles have to be more than 180 degrees.

Since there is the same number of each, the exterior angles must add up to more.

Cinderella was not actually an orphan in any of the early versions of the story.

Her father was alive and living with her. It made it more tragic for poor Aschenputtel  (Cinderella from the Grimm's Brother' version) because her father let his new wife and stepdaughters abuse her.

The Grimm brothers' did not create fairy tales, they collected the ones,  from the German countryside, that had been passed down orally through the generations.  So their versions are very early versions.  They were always very grim, unintended pun there, all the 'bad' people suffered horrible grissly punishments. Much too grissly for the protected ears of children today.   Goodness always triumphed.

There are \(\frac{16!}{3!5!8!}\)different arrangements of the 16 coins. If you fix the 8th and 12th spots as being occupied by by a nickel and a quarter, then you have only 14 coins left to arrange and there are \(\frac{14!}{2!4!8!}\)possibilities for the remaining coints. So the probability we are looking for is\(\frac{\frac{14!}{2!4!8!}}{\frac{16!}{3!5!8!}}\), which I think simplifies to \(\frac{1}{16}\).To see what is happening, let's look at a simpler problem:

Suppose you only have 1 nickel (N), two quarters(Q), and 3 pennies(P). These coins can be arranged in 60 different ways (\(\frac{6!}{3!2!1!}\)).

The following diagram show the sixty different arrangements; of those 12 of them have a quarter in the 3rd and a penny in the 5th spots. so the probability of drawing a quarter on the third and a penny on the 5th draw is \(\frac{12}{60}=\frac{1}{5}\).

Well,
360 = (n - 2)180

2 = n - 2

4 = n
therefore it should be a square.
Correct me if I'm wrong.