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One way to find the fraction that equals  0.77666666666...

Let  x  =  0.77666666666...

To handle the 0.77, multiply this equation by 100:  (because there are two places)

--->     100x  =  77.66666666...

To handle the repeating 6s, multiply this equation by 10:  (because there is one place that repeats)

--->    1000x  =  776.666666...

Subtract these equations:  1000x  =  776.6666666...

                                    -        100x  =    77.6666666...

To get:                                   900x  =  699

Solve this for x:                           x  =  699 / 900                (this can be reduced)

To calculate the average, 

first, for each row, multiply the number of families times the number of cars that they own:

   2 · 0  =  0

   8 · 1  =  8

   x · 2  =  2x

   2 · 3  =  6

second:  add the totals from each row:  0 + 8 + 2x + 6  =  2x + 14

third:  find the total number of families:  2 + 8 + x + 2  =  x + 12

fourth:  find the average by dividing the result of the second step by the result of the third step:

                       average  =  (2x + 14) / (x + 12)

fifth:  set this average to the known average:  (2x + 14) / (x + 12)  =  4/3

sixth:  solve this equation for x; this willbe your answer.

(a² + 2a)x² + (3a + 5)x + 2 = 0

For imaginary solutions, b² - 4ac < 0

(3a + 5)² - 4(a² + 2a)2 < 0

9a² + 30a + 25 - 8a² - 16a < 0

a² + 14a + 25 < 0 => \(a = {-7 \pm 2sqrt{6} }\)

 \(a = {-14 \pm \sqrt{96} \over 2}\)

Hello proyaop!

\(\frac{900^o}{360^o}=2.5\ \to 2\cdot 2=\color{blue}4\)

proyaop, the negative sine values are in the third and fourth quadrants!

4 values satisfy sin x = -0.31.

  !

c = 4d

b = 2c = 8d

a = 3b = 24d

(a + c)/(b + d) = (28d)/(9d) = 28/9

x² + 12x + k = 18x + 3

x² - 6x + (k-3) = 0

Let the roots be a, 2a

a + 2a = 6, a(2a) = k-3 => 3a = 6 => a = 2

2a² = k - 3 => k - 3 = 8 => k = 11.

Thanks for double checking, also I think you can use AC as the diameter of a circle, then knowing ABC is a right triangle and then using the pythagorean theorem, thanks anyway :D

The original expression is:

\(\sqrt{2} + \sqrt{3} + {1\over2\sqrt{2} + 3\sqrt{3}}\)

Next we can simplify the fraction part of the expression by multiplying the denominator by the difference of squares;

The expression is now:

\(\sqrt{2}+\sqrt{3} + {3\sqrt{3}-2\sqrt{2}\over19}\)

Then we can multiply the first two terms of the expression by \(19\) to get:

\({19\sqrt{2}\over19} + {19\sqrt{3}\over19} + {3\sqrt{3}-2\sqrt{2}\over19}\)

Then adding up the numerators and simplifying we get:

\(17\sqrt{2} + 22\sqrt{3}\over19\)

Thus, \(a\), \(b\), and \(c\) are \(17\), \(22\), and \(19\) respectively. 

Since we are looking for the sum of our three "variables", we add \(17+22+19\) to get:

\(58\), which is the answer.

:D

Hi proyaop, yes your answer is correct.

Here is another method, using the cosine law twice only.

Consider triangle ABM, AB=5, AM=4 and BM = 4 (Because the median bisects the opposite side to the vertex).

So applying cosine rule gives you the angle ABC (which is the same as ABM)

Then, consider the triangle ABC, we have, AB=5 and BC=8 and we have the angle ABC.

Then apply cosine rule, yielding AC to be \(\sqrt{39}\)

I got \(\sqrt{39}\) with long equations and polynomials, can someone double check thanks!

The possibilites are:  HHT, HTH, THH, HHH

=  (1/2) x (1/2) x (1/2) = 1/8 for each of the four possibilities.

Adding together these possibilities:  (1/8) + (1/8) + (1/8) + (1/8)  =  ....

If all three letters must be different, 

there are 15 choices for the first letter,

14 choices for the second letter, and

13 choices for the third letter.

So, the total number of choices is:  15 x 14 x 13 = ...

\(\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{8}}+\frac{5}{\sqrt{32}}\)

\(= \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{3}{2\sqrt{2}} + \frac{5}{4\sqrt{2}}\)

\(= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{3}{2 \sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{5}{4\sqrt{2}}\)

\(= \frac{\sqrt{2}}{2} + \frac{3 \sqrt{2}}{4} + \frac{5 \sqrt{2}}{8}\)

\(= \frac{4 \sqrt{2} + 6\sqrt{2} + 5\sqrt{2}}{8}\)

\(=\boxed{\frac{15\sqrt{2}}{8}}\)

We are not here to do your homework for you.

We are not here to do your homework for you.

1 - 0.87 = 0.13

Find the discriminant (b^2 - 4ac) and see which equation has a discriminant of 0.

4x^2 + 16x + 8 ---> 16^2 - 4(4)(8) = 128

-x^2 + 4x + 5 ---> 4^2 - 4(-1)(5) = 36

9x^2 - 6x + 1 ---> (-6)^2 - 4(9)(1) = 0

We know that only one quadratic has one distinct root, so we can stop checking here.

9x^2 - 6x + 1 = 0

(3x - 1)^2 = 0

3x - 1 = 0

x = 1/3

This question was already answered: https://web2.0calc.com/questions/need-help-with-algebra_18

x^2 + (k - 10) + 36

square of a binomial = (x + b)^2 = x^2 + 2bx + b^2

The constant term is 36, so b is 6 or -6.

k - 10 = 12 or k - 10 = -12

k = 22 or k = -2

We can write everything in terms of x, the number of pages Gerald has reawd.

Gerald = x

Nash = x + 148

Josh = 2(x + 148)

x + (x + 148) + 2(x + 148) = 2724

x + x + 148 + 2x + 296 = 2724

4x + 444 = 2724

4x = 2280

x = 570

Gerald has read 570 pages.

subtract the probability that 0, 1, or 2 children don't recognize their mother's voice from 1

\(1 - (0.9)^{20} - (0.1)^1(0.9)^{19} - (0.1)^2(0.9)^{18} \approx \boxed{0.8634}\)

400,000 - 2,345 = 397,655

answers to a) to c) here: https://web2.0calc.com/questions/a-2-2-2-1-b-5-1-5-0-3-1-c-3-10-2-5#r1

\(D) \qquad \frac{6^2 - (-7)^2}{3^{-2} + 3^{-1}} = \frac{-13}{\frac{4}{9}}= \boxed{-\frac{117}{4}}\)

\(A) \qquad (-2)^2 - 2^1=4-2=\boxed{2}\)

\(B) \qquad 5^{-1} + 5^0 - 3 - 1 = \frac{1}{5}+1-4=\boxed{-2.8}\)

\(C) \qquad (3.10)^{-2} + 5 = \frac{1}{(3.10)^2}+5 \approx \boxed{5.104}\)