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angle DAC = 81 degrees.

There are 6 solutions.

Could you please explain how you got 28 so i can understand better?

a + b = 12.

\(\text{Add the two equations together to get }\)\(4(x+y+z)=2400, \text{thus } x+y+z=600.\)\( \text{ Because we have 3 numbers, we divide}\)\(\text{this by 3 to get the average:} \frac{x+y+z}{3}=\boxed{200.}\)

(1881, 3553, 5225, 8778) >>Total = 4 such palindromes.

Using the distance formula,\(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2},\) we substitue them in to get \(\sqrt{(2-(-4))^2+(-6-3)^2},\) which simplifies to \(\sqrt{36+81}=\sqrt{117}=\boxed{3\sqrt13.}\)

a = 14.

The value of n is 28.

The 75 cancels with the -75. The 72 cancels with the -72. Since the difference between the two numbers is 3, and 3|75, all the numbers cancel out and sum to 0.

Sorry, it's supposed to be 5/2% oops

Thanks for the help!

First find the value of \(x.\) Expanding and combining like terms, we find that it becomes a quadratic equation, \(5x^2+6x-18=0\).

Solving using the quadratic formula, we see that \(x\) has two solutions- \(\frac{-3\pm3\sqrt{11}}{5}\). We want the largest possible solution, so the \(\pm\) becomes a \(+\) instead.

Thus we have \(\frac{-3+3\sqrt{11}}{5}=\frac{a+b\sqrt{c}}{d}\). Substituting these values in for each other, we find that \(a=-3, b=3, c=11, d=5.\) So the value of \(\frac{acd}b=\frac{-3\cdot11\cdot5}{3}=\boxed{-55}.\)

Do you have any take on this yet? Have you tried this problem on your own? Since it's marked homework, you should try that first.

\(\frac{343}{3}=114 \frac13\). What did they do to the dogs?!?!

Question-wise, it's \(114 \frac13 \cdot 2 = \boxed{228 \frac23 \text{non-dogs.}}\) 

ITS sqrt(2)/2

2+sqrt(4)

 Angle \(\text{ETF}\) is \(105\)º

I think this is only half of the problem! 

However, researching, I found this problem:

N is a four-digit positive integer. Dividing N by 9 , the remainder is 5. Dividing N by 7, the remainder is 3. Dividing N by 5, the remainder is 4. What is the smallest possible value of  N?

thanks turns out i was wrong