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We know  angle F ,  the adjacent  side  and we  are looking  for  the  hypotenuse......we need  the  cosine

cos F   = EF / DF

cos 60  =  6 / DF

1/2  =  6  /DF

DF =  6 /  (1/2)     =   12

Here is the majority of this answer....you can finish it

After two years :

Jose has     50 000 (1+ .04)²

Patric  has   50 000 ( 1 + .04/2)^(2*2)

I'd love to explain. 

Let's start with an even square. 

On each diagonal, we're starting with a corner, then we're moving 1 row down and 1 column down.

That means the number of squares on the diagonal is the number of rows/columns.

As you can see in the above example, a 4 by 4 square has 4 yellow tiles in each diagonal.

The first tile on the (first row first column). 

The second tile on the (second row second column).

The third tile on the (third row third column).

The forth tile on the (forth row forth column).

Also notice that there are 2 diagonals, with each diagonal having the same number of squares as the row/column. 

So a e by e square has 2e yellow squares.

Now let's look at an odd square. 

The same rule as before applies, where each diagonal has the same number as the number of rows/columns. 

There are also 2 diagonals, but look at the center. 

That yellow tile is shared by both diagonals, meaning we would've counted it twice in the equation 2e. 

So we have to subtract 1 from it. 

2e - 1

I hope this was helpful. :))

=^._.^=

https://web2.0calc.com/questions/counting-problem_45

(a+b)^2 = a^2 + b^2 + 2ab

(a+b)^2 = 40 + 20

(a+b)^2 = 60

(a+b)^2^2 = 3600

=^._.^=

If log3 𝑥 = 𝑡, use the law of logarithms to show that log3 √27 𝑥 is equivalent to 𝑡 + 3 2 .

                               log₃ sqrt 27 x =         =    log₃ x    +   log₃sqrt 27

                                                               =   t   + log₃   27 ^(1/2)

                                                               =  t +    log₃ ( 3³)^1/2

                                                                                     =  t +    log₃ 3^3/2

                                                                                    =  t + 3/2

Your question starts with resistors....and finishes with bulbs....

FOUR resistors in series would all have the same current through them

   v = i r

   v = 2 (20/4) = 10 volts drop across each resistor

Note than 3, 4, 5 is a right triangle, meaning it has a right angle. 

Can you take it from there?

=^._.^=

The teacher has written an equation of the form  on the board,
where  is a quadratic, but Heather can't read the linear term.
She can see that the quadratic term is  4x^2 and that the constant is -24.
She asks her neighbor, Noel, what the linear term is.
Noel decides to tease her and just says, ``One of the roots is 3.''

Heather then says, ``Oh, thanks!''

She then correctly writes down the linear term.
What was the linear term? Include entire term, not simply the coefficient.

\(\begin{array}{|rcll|} \hline 4x^2+ax-24&=& 0 \\ \hline \dfrac{-24}{4} &=& r_1*r_2 \quad | \quad r_1 = 3 \\\\ \dfrac{-24}{4} &=& 3*r_2 \\\\ r_2 &=& \dfrac{-24}{12} \\\\ \mathbf{r_2} &=& \mathbf{-2} \\ \hline \dfrac{a}{4} &=& - (r_1+r_2) \\\\ \dfrac{a}{4} &=& - (3-2) \\\\ \mathbf{ a } &=& \mathbf{ -4 } \\ \hline \end{array}\)

\(4x^2-4x-24\)

Hey there, Guest!

So, 5/8 divided by 1/16 is 10.

Therefore Jackson can cut 10 pieces.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Evaluate the sum
\(\dfrac{1}{3} + \dfrac{2}{3^2} +\dfrac{1}{3^3} + \dfrac{2}{3^4} +\dfrac{1}{3^5} + \dfrac{2}{3^6} + \dots\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{3} + \dfrac{2}{3^2} +\dfrac{1}{3^3} + \dfrac{2}{3^4} +\dfrac{1}{3^5} + \dfrac{2}{3^6} + \dots \\\\ &=& \dfrac{1}{3}*\dfrac{3}{3} + \dfrac{2}{3^2} +\dfrac{1}{3^3}*\dfrac{3}{3} + \dfrac{2}{3^4} +\dfrac{1}{3^5}*\dfrac{3}{3} + \dfrac{2}{3^6} + \dots \\\\ &=& \dfrac{5}{3^2} + \dfrac{5}{3^4} + \dfrac{5}{3^6} + \dots \qquad \text{GP:}~ a=\dfrac{5}{3^2},~ r = \dfrac{1}{3^2} \\\\ s&=& \dfrac{5}{3^2} *\left( \dfrac{1}{1-\dfrac{1}{3^2}}\right) \\\\ s&=& \dfrac{5}{9} * \dfrac{9}{8} \\\\ \mathbf{s}&=& \mathbf{\dfrac{5}{8}} \\ \hline \end{array}\)

A: 335g of CO2

There are 3 cases to work out.

RRRRRRRRLL (right and left) in some combination and LLLLLLLLRR (left and right) in and LLLLLRRRRR. 

2*10C2 + 10C5

2*45 + 252 = 342

I hope that's right. :)

=^._.^=

Edit:

It asked for probability, not number of ways, I also missed one of the cases from before. 

2^10 is that total number of ways. 

342/1024  = 171/512

=^._.^=

(x^4+ax^7+bx+c)(x^3+dx^2+e)(x+f)(x^3-8x^2+7x-1),

x^4 * x^3 * x^3   = x⁴⁺³⁺³    

      The constants  a-f  do nothing to change the degree of the polynomial

Doesn't sound right.

1/130 = .0 076923 076923......

 there are 6 digits that repeat after the initial 0

    130/6    is 21 2/3   so

   the 130 th would normally be a   '9'    ( 2/3 position of the string)   but the string is shifted one digit to the right due to the first '0'

        so the 130th digit will be '6'           As Melody found in her answer ........

If K gave away half and winds up with 9 fewer, then 1/2 of his stickers is 9    ....so K started with 18

   I think your question may be worded improperly....please proofread....

12x+4 = 16    Subtract 4 from both sides of the equation

12x = 12          now divide by 12

x = 1

Oops!

r = 1 / cos(30º) - 1

r = 1 / cos(30º)

1/13 =0.0769230769230769230769230769230769... 

\(1/13 =0.\dot 07692307692\dot3\\ \)

there are 12 repeating digits after the decimal point

130/12= 10 and 10/12

the 10 digit is 9

when you divide this number by 10 the 130th place after the decimal point  will be 6

You need to check this.

V = 180pi cm³

There are two pairs \((x,y)\) of real numbers that satisfy the equation \(x + y = 4xy = 5\).
Given that the solutions x are in the form \(x = \dfrac{a \pm b \sqrt{c}}{d}\) ,
where a, b, c, and d are positive integers and the expression is completely simplified,
what is the value of \(a + b + c + d\)?

My attempt:

\(\text{Let $a=x+y=5$ } \\ \text{Let $b=xy=\dfrac{5}{4}$ }\)

\(\begin{array}{|rcll|} \hline x^2 - Ax + B &=& 0 \\ x^2 - 5x + \dfrac{5}{4} &=& 0 \\\\ x &=& \dfrac{ 5 \pm \sqrt{25-4\dfrac{5}{4}} }{2} \\\\ x &=& \dfrac{5 \pm \sqrt{25-5} }{2} \\\ x &=& \dfrac{5 \pm \sqrt{20} }{2} \\\\ \mathbf{x} &=& \mathbf{\dfrac{5 \pm 2\sqrt{5} }{2}} \\ \hline x+y &=& 5 \\ y &=& 5-x \\ y &=& 5-\dfrac{5 \pm 2\sqrt{5} }{2} \\\ y &=& \dfrac{10-5 \mp 2\sqrt{5} }{2} \\\\ \mathbf{y} &=& \mathbf{\dfrac{5 \mp 2\sqrt{5} }{2}} \\ \hline \end{array}\)

Two pairs (x,y):
\(\begin{array}{|rcll|} \hline (x_1,~y_1) &=& \left(\dfrac{5 + 2\sqrt{5} }{2},~\dfrac{5 - 2\sqrt{5} }{2} \right) \\\\ (x_2,~y_2) &=& \left(\dfrac{5 - 2\sqrt{5} }{2},~\dfrac{5 + 2\sqrt{5} }{2} \right) \\ \hline \end{array}\)

\(a=5,~ b=2,~ c=5,~ d=2 \\ \mathbf{a+b+c+d = 14}\)

If you were to pick two students out of a class of 7 girls and 6 boys,

what is the probability that the students chosen are two boys?

\(\begin{array}{|rcll|} \hline \dfrac{ \binom{6}{2}_{\text{Boys}}\binom{7}{0}_{\text{Girls} } } {\binom{13}{2}_{\text{Boys and Girls} } } &=& \dfrac{5}{26}\\ \text{or} \\ \dfrac{6}{13}*\dfrac{5}{12}&=& \dfrac{5}{26}\\ \hline \end{array}\)

The probability is \(19.23 \%\)