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Correct! :D

The actual roots are 2 and -5.

By Vieta's formulas, a + b = 5 and ab = -11, so (a + 3)(b + 3) = ab + 3(a + b) + 9 = -11 + 3*5 + 9 = 13.

It's fine. Do you know what the real answer is though?

-x^2 + 18x + 81 = -(x - 9)^2, so this is the quadratic with a repeated root.

Thank you!

Then the final answer is 168/6 * 6/35 = 1008/210

= 24/5

So we have a number \(x\) that has this property that makes the following equation true:

\(x^2 = 84 + 5x\)

Changing this to standard quadratic form, we get:

\(x^2 - 5x - 84 = 0\)

This, factored into two binomials, is:

\((x + 7)(x - 12) = 0\) 

From this, \(x\) must be either \(\fbox{$-7 \text{ or 12}$}\). I guess those are your numbers :D

AB = 7*sqrt(10).

Can you attach a diagram?

We can start by finding squares of different sizes:

Squares of side length 1 (small): 16

Squares of side length 1 (normal): 19

Squares of side length 2 (normal): 9

Squares of size length 3 (normal): 4

Squares of size length 4 (normal) 1 (the entire thing! :D)

The total would be 16 + 19 + 9 + 4 + 1 = 49 squares :D

Fusion Power

"One of the biggest reasons why we haven't been able to harness power from fusion is that its energy requirements are unbelievably, terribly high. In order for fusion to occur, you need a temperature of at least 100,000,000 degrees Celsius. That's slightly more than 6 times the temperature of the Sun's core." 

Quoted from https://www.scienceabc.com/eyeopeners/why-arent-we-using-nuclear-fusion-to-generate-power-yet.html :D

You're right -- I messed up -- I thought (3,2) was (2,3) -- my fault ...

Hmm. This means to represent \(\dfrac{28}{5 \frac{5}{6}}\) as a fraction. We could turn them both into improper fractions:

\(28 = \frac{168}{6}\)

\(5 \frac{5}{6} = \frac{35}{6}\)

Dividing \(28\) by \(5 \frac{5}{6}\):

\(\frac{168}{6} \div \frac{35}{6}\)

When dividing one fraction by the other, you simply multiply by the right fraction's reciprocal:

\(\frac{168}{6} \div \frac{35}{6} = \frac{168}{6} \cdot \frac{6}{35}\)

So we just have to multiply these fractions. What do you get? :D

That was incorrect. Thank you for trying though. Any idea on what went wrong?

Please fix the LaTeX.

A  =  (-2, 0)

B  =  (2, 0)

C  =  (3, 2)

Since there is a right angle at B, AC must be a diameter of the circle.

Therefore, the center of the circle is the midpoint of AC.

The midpoint of AC  =  ( (-2 + 3)/2, (0 + 2)/2 )  =  ( 1/2, 1 )

Wow!!!....thanks, geno....I've never seen that "formula" before   !!!

There is a rather obscure formula for the sum of two inverse tangents:

tan^-1(a) + tan^-1(b)  =  tan^-1(  (a + b / ( 1 - a·b ) )

tan^-1( 1/2 ) + tan^-1( 1/3 )  =  tan^-1(  ( 1/2 + 1/3 ) / ( 1 - ( 1/2 )·( 1/3 ) )

                                        =  tan^-1( (5/6) / (5/6) )

                                        =  tan^-1( 1 )

                                        =  pi/4

Adding this to  tan^-1( 1 )  gives a total of  pi/2  [or 90^o]

See the answers here  :

https://web2.0calc.com/questions/rectangle-abcd-contains-a-point-such-that-ax-1-bx

alpha     =  tan^-1(1/1)

beta       =  tan^-1(1/2)

gamma  =  tan^-1(1/3)  

The sum will be  pi/2.

It's  just really computing the areas of 6 right triangles and subtracting these combined areas from the area of the square

First right triangle  ABE

The legs are BE  = 6     AB  =12

Area  = (1/2) prduct of the legs  = (1/2)(6)(12) = 36

We also  need to find AE  =sqrt  [ BE^2 + AB^2]  =sqrt [ 6^2 + 12^2] = sqrt 180  =  6sqrt (5)

Next  right triangle FCE   ..... [FC  = DC/3 = 4]

Area  1/2( CE)(FC)  = (1/2)(6)(4)   =12

And  FE  = sqrt [ CE^2 + FC^2]  =sqrt [ 6^2 + 4^2]  = sqrt 52  = 2sqrt (13)

Next triangle FDG...DF = 8  and  DG   = 3

Area= (1/2) (8)(3)  = 12

GF  =sqrt ( 8^2 + 3^2)  = sqrt [73]

Next triangle GAH  ....AH  = AE/ 3  = 2sqrt 5  =  sqrt (20)

We need to find GH  =  sqrt  [GA^2  - AH^2]  =  sqrt [ 9^2 - 20]  = sqrt (61)

Area of GAH  =(1/2) (AH)(GH)  = (1/2)(sqrt (20) (sqrt(61)  = sqrt (5)sqrt (61) = sqrt (305)

Area of FJG ...we know FJ  =(1/2)FE  = sqrt (13)

Area=  (1/2)(GF)(FJ)  =  (1/2)sqrt (73)sqrt (13)  = (1/2)sqrt (949)

Lastly  area of  JEH  ....JE  = sqrt (13)    HE =  (2/3)AE  = 4 sqrt (5)

Area  = (1/2)(JE)(HE) =  (1/2)sqrt (13) 4sqrt (5)  = 2 sqrt (65)

So....area  of  purple triangle  =

12^2  -  [ 36 + 12 + 12 + sqrt (305) + sqrt (949)/2 + 2sqrt (65) ]  ≈  35  units^2

If I didn't make any errors   !!!!

sum  =  tan^–1(1/1) + tan^–1(1/2) + tan^–1(1/3)  

sum  =  45.000^o + 25.565^o + 18.435^o  

sum  =  89.000^o 

_.

Given a rhombus with diagonals  14  and  48  --  what is the perimeter?

Since the diagonals of a rhombus are perpendicular to each other, four congruent right triangles are formed.

Each of these triangles have bases of  7  and  24  so we can use the Pythagorean Theorem to find the side

of the rhombus.

c2  =  72 + 242     --->     c2  =  49 + 576     --->     c2  =  625     --->     c  =  25.

The perimeter will be  4 x 25  =  100

For your second question:  https://web2.0calc.com/questions/quadrilateral-problem_1

Given a rhombus with diagonals  14  and  48  --  what is the perimeter?

Since the diagonals of a rhombus are perpendicular to each other, four congruent right triangles are formed.

Each of these triangles have bases of  7  and  24  so we can use the Pythagorean Theorem to find the side

of the rhombus.

c²  =  7² + 24²     --->     c²  =  49 + 576     --->     c²  =  625     --->     c  =  25.

The perimeter will be  4 x 25  =  100