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The number has to be even, so there are 2 cases: the final digit is a 2, or the final digit is a 4. 

If the final digit is a 2:

3 choices for the first number (can't be 2 or 5)

3 choices for the next number (can't be the number just chosen or a 2)

1 choice for the final number (must be a 2)

We can do the same thing for the other case (the final digit is a 4), so there are \(3 \times 3 \times 2 = \color{brown}\boxed{18}\) numbers that work. 

\(22=-t^2+9t+2\)

\(t^2 - 9t + 20 = 0\)

\(-4 \cdot -5 = 20, -4-5=-9\)

\((t-4)(t-5) = 0\)

\(t - 4 = 0\)

\(t = 4\)

\(t - 5 = 0\)

\(t = 5\)

The rock will reach a height higher than 22 meters \(\boxed{4}\) seconds after it's released. It will stay above 22 meters for \(5 - 4 = \boxed{1}\) seconds.

Thank you for your explanation, yes for several months I forgot this website's name until I remembered today lol. yes I inputed it wrong, ill put it into my next post thanks again！

Hi Jenny,

It is nice to see you back again.

No.2

\(log_3(x^2-22)=3\\ 3^{log_3(x^2-22)}=3^3\\ x^2-22=27\\ x^2=49\\ x=\pm7 \)

Can you please check the the first one is inputed correctly. 

And next time put them on a seperate posts.  (1 post = 1 question)

There are 6 choices for the shirt,

there are 3 choices for the pants,

there are 2 choices for the shoes,

and there are 5 choices for the tie (4 ties plus no tie)

for a total of 6 x 3 x 2 x 5 choices for the outfit.

There are 4 choices for the left-most digit (1, 2, 3, or 4; 5 can't be chosen)

then there are 4 choices for the middle digit and 3 choices for the right-most digit.

This means that there are 4 x 4 x 3 = 48 choices.

Ok, 13 units     

Here is the great circle of the sphere  (The equator if you like)

CA will be the longest possible distance, in a straight line from A to C

Method:

1) Use the cosine rul to find angle alpha (keep it exact)

2) Find DX (exact)

3) find XC Pythagoaras's theorem (exact)

4) CA = 2* CX

I have not an arc distance, it is the straight distance.)

If you want the longest possible arc distance then you can find it afterwards easily enough. 

The radius of the sphere is 13, not 3

Let the mass of sugar in container B==M
The mass of sugar in container A ==1/4M

[M - 14.4] / [1/4M - 2.25] ==2, solve for M

M ==99/5 ==19.80 kg - mass of the sugar in container B at first.
1/4 * 19.80 ==4.95 kg - mass of the sugar in container A at first.

Welcome to the forum,

but please next time proof read your question and get rid of meaningless dollar signs.

Points A,B and C are on a sphere whose radius is 3. 

If AB=BC=10 what is the longest possible value of AC?

No 2 points on the sphere can be more than 6 units apart. (in a straight line). 

If you are talking the arc distance then the greatest distance is half of  9pi   (half the circumference of the great circle)

4.5pi is approx 14.13 units.

I would think that the points could be at most 4.5pi  units apart.

Find the sum of all possible values of x.

Hello Guest!

\(y=\sqrt{(16\sqrt{2})^2-(x-4)^2}-7\\ y=21\\ 21=\sqrt{512-(4-x)^2}-7\\ 28^2=512-16+8x-x^2\)

\(x^2-8x+286=0\\ x=4\pm\sqrt{16-286}\\ x=4\pm 3i\sqrt{30}\)

\(x\ne\mathbb R\) | No solution.

  !

Simplify the equation to: \(x^2-12x+y^2+14y=36\)

Complete the square for the left hand side:\((x-6)^2+(y-7)^2\)

Simplifying this gives us: \(x^2-12+y^2+14y+85\)

Since we added 85 to the left hand side, we have to add 85 to the right hand side as well.

This gives us: \((x-6)^2+(y+7)^2=121\)

The right-hand side is the radius squared, so we have the equation: \(121 = r^2\)

Solving, we find \(\color{brown}\boxed{r = 11} \)

Marking the price up by 100% means the fountain costs \($1,991.32\)

Marking it down 25% meanst the fountain costs \(3 \over 4\) the new price, meaning it costs \(0.75 \times 1991.32 = \text{____}\)

This problem isn't possible. The shortest possible line segment would be between \((4, -7)\) and \((4, 21)\).

This line segment has a length of 28, which is greater than \(16 \sqrt2\), so there are no points that work. 

how?

Draw line \(AM\) so it is perpendicular to \(DC\)

Trapezoid \(ABCD\) is isceles, meaning \(DM = 1\)

Applying the Pythagorean Theoram to \(\triangle AMD\), we find that \(AM = 1\)

Applying the Pythagorean Theorem to \(\triangle AMC\), we find \(\color{brown}\boxed{x=\sqrt{5}}\)