All Answers 358512 Answers

The area is 40.

Angle CBD = 59 degrees.

The shaded area is 36 - 6*sqrt(3) - 8*pi, so the answer is 36 + 6 + 3 + 8 = 53.

sin angle ABC = 2/7.

The radius of the circle is 16/3.

The area of triangle BMD is 140.

The answer is 240.

I think that this would mean that when you plug 3a-2 into this equation you would get 1 so that would mean 1=(2x-3)(x+1) Expand to 2x²-x-3=1 set equal to 0 so 2x²-x-4=0 and then use the quadratic equation (1+-sqrt(1+32))/4 so 3a-2=(1+-sqrt(33))/4

Combine like terms: x^2 + 10x + 25 = 25, subtract 25 from bothsides, x^2 + 10x = 0, you'll see that it can be factored to x(x + 10) = 0, so either of those are equal to 0, if looking for the smaller one, look at x + 10 = 0, x = -10.

-b / 2a = 2

b = -4a

ax^2 - 4ax + c.

a - 4a + c = -1

-3a + c = -1

plug in 2

4a - 8a + c = 4

-4a + c = 4

Subtract equation 2 from equation 1

a = -5

c = -16

b = 20

-5x^2 + 20x - 16

Attn  Keihaku

I am waiting for a response from you.

P==Purple,  R == Red,  G ==Green

P / R == 1/9,   

R / G == 3 / 5,  solve for P, G 

P == R / 9

G == 5R / 3

P / G == [R / 9]  / [5R / 3] == 1 / 15

The measure of ∠FHJ is 60°. The measure of ∠FGH is 25°. What is the measure of m∠HFG? Enter your answer in the box.  

A straignt line is 180^o so FHG is (180^o – 60^o) = 120^o  

The sum of the interior angles of a triangle is 180^o so HFG is (180^o – 120^o – 25^o) = 35^o  

_.

Fill in the blank to make the equation true.  

13*5*13*___*3 = 13*17  

Call the blank "x". 

Repeat the equation as             13 • 5 • 13 • x • 3  =  13 • 17  

Divide both sides by 13                     5 • 13 • x • 3  =  17  

Divide both sides by 5 • 13 • 3                             x  =  (17 / 195)    this won't reduce any further  

_.

How many times have you submitted this question in the last week or two ?? What is the reason that you don't get your question answered? In my opinion, it is because nobody knows how to answer it, because not many are good at answering questions in Statistics and Sets and Subsets and Elements of a Set....etc.

Good luck to you.

Let DEF be an equilateral triangle with side length 3. At random, a point G is chosen inside the triangle. Compute the probability that the length DG is less than or equal to 1.  

Draw equilateral triangle DEF with side length 3.  

Place the point of a compass on D and draw a circle with radius 1. 

Any point inside the piece of the triangle that's set off by the circle's arc will be 1 or less from D. 

Your probability is the area of the piece set off by the arc divided by the area of the entire triangle. 

Figure the area of the circle, π r²  

then take a sixth of it for the area of the segment.    (60/360) • 3.14 • 1²   

Figure the area of the triangle, (1/2) bh. 

You can get the height with Pythagoras' Theorem.    h = sqrt(3² – 1.5²)  

My answer:   (0.524 / 3.897) = 0.135  or as a percentage  13.5% probability   

_.

How often do you check to see if your question is answered or not ?? Did you check under your last post here ?

https://web2.0calc.com/questions/counting-problem_29576

What "iterative formula" ??!

Draw up a Venn diagram and upload it. Don't expect many people to draw you a Venn diagram and solve your problem for you. That is the reason you don't get any answer, in my opinion. Good luck.

Point C

Hello Guest!

\({\color{blue}m}=\frac{y_B-y_A}{x_B-x_A}=\frac{8-7}{6-2}=\color{blue}\frac{1}{4}\\ y=m(x-x_A)+y_A=\frac{1}{4}(x-2)+7\\ \color{blue}y=\frac{1}{4}x+7.5\)

\(\Delta x=\frac{x_B-x_A}{6}=\frac{6-2}{6}=\frac{2}{3}\\ {\color{blue}x_C=}\ x_A+\Delta x=\color{blue}2\frac{2}{3}\\ y_C=\frac{1}{4}(x-2)+7\\ \color{blue}y_C=7.1\overline 6=\frac{43}{6}\)

Point \(C\ (2\frac{2}{3},7\frac{1}{6})\) is on the segment connecting A and B such that the distance from the point C to B is 5 times the distance from the point C to A.

  !

Probability of exactly one six:        p1 = (1/6)*(5/6)^2*3

Probability of exactly two sixes:   p2 = (1/6)^2*(5/6)*3

Probwbility of exactly three sixes:  p3 = (1/6)^3

Expected number of sixes:  1*p1 + 2*p2 + 3*p3

Opposite angles in a cyclic quad are suplementar

Givew those equalities I can see that

The smallest would add with the biggest to get 180 degrees

The largest angle could be

\(\frac{\angle q}{4}=\frac{\angle r}{5}\\ \angle q=0.8*\angle r\\~\\ \angle q + \angle r =180^ \circ\\ 0.8*\angle r + \angle r =180^ \circ\\ 1.8*\angle r =180^ \circ\\ \angle r =100^ \circ\\ \angle q = 80 ^\circ\\~\\ \frac{\angle p}{3}=\frac{\angle r}{5}\\ \frac{\angle p}{3}=\frac{100}{5}\\ \angle p = 60^\circ\\ so\\ \angle s = 120 ^\circ \)

LaTex:

\frac{\angle q}{4}=\frac{\angle r}{5}\\
\angle q=0.8*\angle r\\~\\
\angle q + \angle r =180^ \circ\\
0.8*\angle r + \angle r =180^ \circ\\
1.8*\angle r =180^ \circ\\
\angle r =100^ \circ\\
\angle q = 80 ^\circ\\~\\
\frac{\angle p}{3}=\frac{\angle r}{5}\\
\frac{\angle p}{3}=\frac{100}{5}\\
\angle p = 60^\circ\\
so\\
\angle s = 120 ^\circ