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61, 61, 61, 178, 179

This is not math...

In a certain pentagon, the interior angles are   A B C D    and E   where  A B C D E  are integers strictly less than 180 ("Strictly less than 180" means they are "less than and not equal to" 180.)
If the median of the interior angles is  61  and there is only one mode,

then what are the degree measures of all five angles?

Given that  $\large{x}$  is a positive integer less than 100,
how many solutions does the congruence  have?
$\large{x + 13 \equiv 55 \pmod{34}}$

$\begin{array}{|rcll|} \hline x + 13 &\equiv& 55 \pmod{34} \\ x + 13 &\equiv& 55-34 \pmod{34} \\ x + 13 &\equiv& 21 \pmod{34} \quad & | \quad - 13 \\ x &\equiv& 21-13 \pmod{34} \\ x &\equiv& 8 \pmod{34} \\ \mathbf{x} & \mathbf{=} & \mathbf{8 + n \cdot 34}, ~ n \in \mathbb{N} \\ \hline \end{array}$

$\begin{array}{|c|l|c|} \hline n, ~ n \in \mathbb{N} & \mathbf{x = 8 + n \cdot 34} & ~ x>0,~ x<100 \\ \hline 0 & x = 8+0\cdot 34 \\ & x= 8 & \checkmark \\ \hline 1 & x = 8+1\cdot 34 \\ & x= 42 & \checkmark \\ \hline 2 & x = 8+2\cdot34 \\ & x = 76 & \checkmark \\ \hline 3 & x = 8+3\cdot 34 \\ & x = 110 & x> 100 \\ \hline \ldots & & x> 100 \\ \hline \end{array}$

The congruence  has three solutions: $\mathbf{x = 8}$ and $\mathbf{x = 42}$ and $\mathbf{x = 76}$

$|M|=2|T|\\ |M\cup T| = |M|+|T|-|M\cap T|\\ |M\cup T|=2x+x - y = \\ 3x-y$

(61,61,61,178,179)

is the only combo that does the job.

$\text{well the largest 4 base 3 digit palindrome is }\\ p = 2222 = 80d\\ \text{I don't see anything more clever than listing this in a few bases until we find a palindrome}\\ \text{In order to have 3 digits it must be that }b^2 < 80d < b^3 \text{ so }\\ 5 \leq b \leq 8\\ \left( \begin{array}{cc} 5 & \{3,1,0\} \\ 6 & \{2,1,2\} \\ 7 & \{1,4,3\} \\ 8 & \{1,2,0\} \\ \end{array} \right)\\ \text{and we see that base 6 is the only base that produces a 3 digit palindrome of }80d$

$\text{let }N \text{ be the number of planets in the solar system}\\ N \pmod{13} = 6\\ N \pmod{14}=5 \\ \text{13 and 14 have no common factors so any solutions will be of the form }\\ x= r + 13\cdot 14 k = r+182k,~r,k \in \mathbb{N}\\ \text{we can eyeball that 19 is a solution and so all solutions will be }\\ N = 19 + 182k \text{, and the first of these such that N > 100 is }\\ N = 19+182 = 201$

$P[A\cup B] = P[A]+P[B] - P[A \cap B]\\ A= \{\text{those who like to read historical fiction}\}\\ B = \{\text{those who like to read autobiographies}\}\\ P[A\cup B] = 0.47 + 0.42 - 0.14 = 0.75$

Put the 0.75 in the box... (or it gets the hose)

See the answer here:  https://web2.0calc.com/questions/help_53845

x^3 + (p+x)x^2 +2px = 0

$x^3 + (p+x)x^2 +2px = 0\\ x[x^2 + (p+x)x +2p]= 0\\ \text{x =0 is one solution}\\~\\ \text{If x is not 0 then}\\ x^2 + (p+x)x +2p=0\\ x^2 + x^2+px +2p=0\\ 2x^2 +px +2p=0\\ x=\frac{-p\pm\sqrt{p^2-4*2*2p}}{4}\\ x=\frac{-p\pm\sqrt{p^2-16p}}{4}\\~\\ so\\ x=\frac{-p\pm\sqrt{p^2-16p}}{4}\qquad or \qquad x=0$

Solve for x:
2 p x + x^3 + x^2 (p + x) = 0

Expand out terms of the left hand side:
2 p x + p x^2 + 2 x^3 = 0

Factor x from the left hand side:
x (2 p + p x + 2 x^2) = 0

Split into two equations:
x = 0 or 2 p + p x + 2 x^2 = 0

Divide both sides by 2:
x = 0 or p + (p x)/2 + x^2 = 0

Subtract p from both sides:
x = 0 or (p x)/2 + x^2 = -p

Add p^2/16 to both sides:
x = 0 or p^2/16 + (p x)/2 + x^2 = p^2/16 - p
Write the left hand side as a square:
x = 0 or (p/4 + x)^2 = p^2/16 - p

Take the square root of both sides:
x = 0 or p/4 + x = sqrt(p^2/16 - p) or p/4 + x = -sqrt(p^2/16 - p)

Subtract p/4 from both sides:
x = 0 or x = sqrt((p^2)/16 - p) - p/4 or p/4 + x = -sqrt(p^2/16 - p)

Subtract p/4 from both sides:

x = 0     or     x = sqrt((p^2)/16 - p) - p/4    or     x = -p/4 - sqrt((p^2)/16 - p)

Putting your question in LaTex, it is $x^3 + (p+x)x^2 + 2px = 0$.

Let's first factor this out. Using the distributive property, we have $x^3 + px^2 + x^3 + 2px = 0 \Rightarrow 2x^3 + px^2 + 2px = 0$.

We want to isolate p, so we subract 2x^3 from both sides to get $px^2 + 2px = -2x^3$. Factoring out p, we have $p(x^2 + 2x) = -2x^3$. Dividing x^2 + 2x from both sides, we have $p = -\dfrac{2x^3}{x^2+2x} \Rightarrow \boxed{p = -\dfrac{2x^2}{x+2}}$.

 - PM

[3x (x - 2)] + [x(x - 4)]=36

4 x^2 - 10 x = 36
2x^2  -  5x  -  18 =0
Use the quadratic formula to find x:
x =4.5                    and x = -2 (discard)
Therefore, the length of AB = x = 4.5 cm

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?

D.   symmetry  

[ the left half and right half are "mirror " images ]

C.  phi   

[ However....the lowercase usually denotes the reciprocal of the  ratio...."Phi" usually denotes the ratio itself ]

3-4v =  -6v+11

Add 6v to both sides

3 + 2v = 11  

Subtract 3 from both sides

2v  =  8       

Divide both sides by 2

v =  4

(2/5) is the scale factor

a.   the ratio is just the scale factor  = 2/5  =  2 : 5

b. the ratio is   (scale factor)^2   =  (2/5)^2 =  4/25   =   4 : 25

c.  the ratio is   (scale factor)^3 =  (2/5)^3  = 8/125  =  8 : 125

One line of symmetry.......a vertical line "splitting" the A into mirror images

Here you go :

https://www.desmos.com/calculator/3mklloavnm

The triangle is isoceles......each of the smaller angles = 30°

To find the straight edge length we can use the Law of Sines

Edge Length / sin120°  = 60/ sin 30°

Edge Length  =   60*sin (120°) / sin 30°  = [60sqrt (3)/2] / (1/2)  = 60sqrt(3) cm   (1) 

For the curved part.....the perimeter is

(240/360) * 2 * pi * (60)  =

(4/3) * pi * 60  =

80 pi  cm    (2)

So....the perimeter is (1) + (2)   =

[ 60sqrt(3) + 80 pi ] cm ≈   355.25 cm