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Altitudes AX and BY of acute triangle ABC intersect at H. If angle BAC = 43 degrees and angle ABC = 67 degrees, then what is angle HCA?

ye i also did the part where u use f(x')= 12 but i dont get where the x-1 come from

It seems to me that the question as stated is incomplete - see my analysis below.

The only way I can see that 16 is the smallest value for abc+d-2 is if a,b,c and d must all be positive integers.

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hi CoolStuff,

It is not really a matter of making more sense, I am not saying my answer is smallest possible but

I think my answer is correct, in that if all these pronumerals have to be integers, and not necessarily positive integers then 

-2 is a solution to that expession.  Obviously -2 is smaller than 16.

Are you sure you have copied the question exactly?

Have you checked for yourself that my answer works?     

Maybe it doesn't, you should check that it meets all the criterion, I could have made an error.

Did they provide any working for their answer of 16?

A rhombus has 4 equal sides.....   148/4 = 37 side length

The diaonals of a rhombus bisect each other (at right angles)....we can find the 1/2 length of the other diagonal

 since 1/2 of the diagonal given and the other 1/2 form a right triangle with one side of the rhombus (37) as a hypotenuse

24/2 = 12     12^2 + x^2 = 37^2      x = 35     so the other diagonal is 2*35= 70

Area Rhombus  = diagonal1 x diagonal2  /  2 =   24 x 70/2 = 840 sq units

I agree so much!

HAHA yes, pretty helpful! I didn't even think of any smarter answers!

what about a=36 and a=-16....those work too

Rom is right!!
Look at the following squares of prime numbers:

2^2= 4
3^2 =9
5^2 =25
7^2 =49
11^2 =121
13^2 =169
17^2 =289 
19^2 =361
23^2 =529
29^2 =841
31^2 =961
37^2 =1,369
41^2 =1,681
43^2 =1,849
As you can see, there are 14 squares of ALL the prime numbers below the sqrt(2010), or 45, that meet your 2 conditions:
1 - They are ALL between 1 and 2010.
2- ALL their divisors sum up to =1 + sqrt(i) + i
3 - As an example take any of the 14 squares above such as 29^2 =841 =1 + sqrt(841) + 841 =871(sum of its divisors)                          

ye i also did the part where u use f(x')= 12 but i dont get where the x-1 come from

That doesn't make a lot of sense. Shouldn't it be the composite odd numbers less than 2010? Because the the number of divisors of the square of a prime number is 3.

Show an equation of the tangent to the curve 𝑦 = 𝑥^5 − 3𝑥^2 +𝑥 + 5 at the point (−1, 0) is given by 𝑦 = 12(𝑥 +1) by differentiation

y' = 5x^4 - 6x + 1

The slope of the tangent line at (-1,0) is

y' (-1) =  5(-1)^4 - 6(-1) + 1      =    12

The equation of the tangent line to (-1,0) is

y = 12 ( x - 1) + 0

y = 12(x - 1)

Suppose we take this problem statement directly at it's word.

\(\text{The sum of the divisors of }n \text{ is }1 + \sqrt{n} + n \text{ i.e.}\\ \text{the divisors are in fact }1,~\sqrt{n},~n\)

\(\text{This implies 2 things}\\ n = p^2 \text{ for some integer }p\\ \text{all of the divisors of }n \text{ are }1, ~p,~p^2\\\)

\(\text{Thus we are looking at the squares of the prime numbers}\)

I'll let you assemble a list of the squares of the primes that are less than 2010.

You might also try to prove that the squares of the primes are the only numbers that satisfy the requirement.

Angle OQL = 90°

Angle OLQ bisects the interior angle MLQ =  120/ 2 =  60°

So...angle LOQ =  180 - 90 - 60 =   30°

CORRECTED......

Thank you so much for answering my question so thoroughly!

The number of diagonals in a regular polygon of n sides = n(n - 3) / 2

So

n(n - 3) / 2 =  5n     multiply boh sides by 2

n(n -3) = 10n       simplify

n^2 - 3n = 10n     subtract 10n from both sides

n^2 - 13n = 0      factor

n ( n - 13) = 0

Set each factor to 0 and solve for n and we get

n= 0   [ reject]    or        n = 13

So....13 sides

See the answer here:  https://web2.0calc.com/questions/geometry_52644

See the answer here: https://web2.0calc.com/questions/geometry_74984

See the answer here: https://web2.0calc.com/questions/a-regular-hexagon-has-perimeter-p-and-area-a-compute-p-2-a_1

Let the side length of the octagon = S.....this = the hypotenuse of an isosceles triangle

So...the length of one of the legs of the isosceles triangles =   S/√2

And the side of the square = 

S + 2S/√2 =  S + S√2

So

S + S√2  = 10

S(1 + √2)  = 10    divide bot sides by ( 1 + √2)

S =     10   / ( 1 + √2 )   ≈    4.14  (inches)

LOL!!!

Actually......your first answer was more "complete" if we allow y to be positive or negative....!!!

Remember that y > 0

(a + bi)^2  = 2( a - bi)

a^2 + 2abi - b^2 =  2a - 2bi

Equating the real and imaginary parts, we have that

(a^2 - b^2) = 2a              2abi = - 2bi

Using the second equation

2a = -2

a = -1

And using the first equation

(-1)^2 - b^2 = -2

-b^2 = -3

b^2 = 3

b = ±3

6 < y < 7   and     -7 < y < -6    according to W/A.