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Do you mean \(1.4e+0.7\) . If then, the answer is \(1.4(2.7)+0.7\approx4.48.\)

Hmm, is it (A), isometry?

I think this is correct....

Based on what we have the area=84 and the semiperimeter=21. Therefore, the incircle(inradius)=84/21=4.

By tangent formulas, we see that the two sides of the quadrilateral are x. So, we have following from the top of the triangle, starting with the side length of 15: x, 15-x, Side length of 14: x-1 and 15-x, Side Length 13: x-1 and x. So, x-1+x=13, 2x-1=13, 2x=14, and x=7. Now, quadrilateral AEIF is made up of two triangles, both with areas of 14, since \(\frac{1}{2}*7*4=14\) . Thus, the answer is \(14+14=\boxed{28}.\)  

Pretty tough one... I would wait for an answer.

I think we should use Heron's here for the area of the triangle, and we have sides 13, 14, and 15.

To use Heron's, we also need to find the semiperimeter, which is \(\frac{13+14+15}{2}=\frac{42}{2}=21.\)

Now, we have \(\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21(8)(7)(6)}=\sqrt{7056}=84.\)

Hmm, I don't know how to continue from here, though...this was my best shot.

Uh, what should we solve here?

cost of an adult ticket: $7.00, let adults be (a)

cost of a student ticket: $5.00. let students be (s)

We have a+s=45 and 7a+5s=265, based on systems of equations.

Therefore, by multiplying by 5 for the first equation, we have 5a+5s=225 and 7a+5s=265.

So, 2a=40, a=20. Thus, the number of students tickets sold were 45-20=25 tickets.

1. We subtract one from each power: 9, 25, 36, 64. Now, taking the prime factorizations, we have \(3^2, 5^2, 2^2*3^2, 2^6\) . By scanning, we can see that 64 is a cube number, therefore 65 is our answer. 

We know that 21 has factors of 1, 3, 7, 21. Remember, when you want to find the number of factors a number has, you just add one to each power, then multiply. Therefore, by picking 7 and 3, we should subtract one from each power, yielding us with 6 and 2. For it to be the smallest integer, we just plug in 2 with the 6 and 3 with the 2. Thus, the answer is \(2^6*3^2=64*9=\boxed{576}.\)

Notice that after 5! every number will end in 0. So, we are just looking for 1!+2!+3!+4! which ends in a 3. 

We set up equations. a+b+c=221, a=b+8, a=c+5