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FHI is the total of FHG and GHI.  And it's given that FHI is 142^o.  

So, set it up:  (3x + 6) + (9x – 8) = 142^o   and then just solve for x and plug that value into each expression. 

Read no farther if you can take it from here. 

Otherwise:                                          (3x + 6) + (9x – 8)  =  142^o  

Add up the like terms                          3x + 9x + 6 – 8  =  142  

                                                                       12x – 2  =  142 

                                                                       12x        =  144  

                                                                                  x  =  12  

Plug in the value into each angle               (3 • 12) + 6  =    42^o  angle FHG  

                                                                   (9 • 12) – 8  =  100^o  angle GHI  

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b - See detailed step by step answer by "heureka" in the middle of page 531 here:  https://web2.0calc.com/members/heureka/?answerpage=531

The number of possible seatings is 6!*5 = 3600.

The probabiity works out to 3/20.

If the angle is   arctan a/b    then it is the angle in the upper left corner because tan = opp/adj

   the cos of that angle is   adj/ hyp =   b / (sqrt(a²+b²)         (because hypot in rt triangle = sqrt(a^2 + b^2)      AND COS = adj / hyp )

??? can you a step by step

The correct answer is D) The student incorrectly wrote the coefficients of the polynomial.

Even though there is no x^2 term presented, when performing synthetic division you need to write in a 0 for the missing terms.

The division should have looked like:

2  I

__I     1     0     2    -3

                 2     4     12

____________________

          1     2     6     9

for XYZ ~ PQR, the ratio of the sides need to be the same.

XY : PQ = 2 : 3

XZ : PR = 12 : 3(x-2)

12 : 3(x-2) = 2 : 3

6(x-2) = 36,

x-2 = 6

x = 8

The only numbers that can be squared to end in 6 are numbers that end in 4 or 6.

We need to find the possible values of the tens digit, so lets let the tens digit be a variable such as x.

The last two digits of our number will look like this:

10x + 6

or this:

10x + 4

I find it easiest to just plug in values from (1-9) for x and see what you get.

If x is 1, the last two digits will be ...56 or ...96

If x is 2, the last two digits will be ...76 or ...76

If x is 3, the last two digits will be ...96 or ...56

If x is 4, the last two digits will be ...16 or ...36

If x is 5, the last two digits will be ...36 or ...16

.

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We can keep going all the way to x=9, and you will see that the tens digit will always be an odd number...

Therefore the possible values of the tens digit is (1, 3, 5 , 7, 9)

can you explain this

The possible values of the tens digit are 1, 2, 4, 5, and 6.

5x + 3 +5x - 5 = 78

x = 8

angle DBC = 5x + 3 = 43º

angle ABC = 5x - 5 = 35º

(4+2i) / (-1+i)

= (4+2i) / (i-1)

(4+2i)(i+1) / (i-1)(i+1)

= 2(i^2)+6i+4 / (i^2)-1

= 6i-2+4 / -1-1

= 6i+2 / -2

= -3i-1

a -  sum_(n=1)^26 (-1)^(n + 1) (105 - 4 n)^2 = 5304

Need more info...or a diagram.....is this in a triangle?   ...or what?

3rd + 6th term = 25        let x = third term

x + 3d = 25

AND    x - 2d= 2         x = 2 + 2d

sub into the first equation

2 + 2d + 3d = 25

5d = 23

d = 23/5

1st + 3d = 4th term

2 + 3 (23/5) = ........

\(\cos 5^\circ = \sin 25^\circ + \sin \theta \)

Hello Guest!

\(\theta=arcsin(cos5°-sin25°)=\color{blue}35°\)

  !

IF your equation is correct    just put   0 in for height ('y')    and solve the equation for 't'   (likely via the quadratic formula  throw out any negative 't' values found)

Given   density    and volume       mass = density * volume   (make sure they are all in the same UNITS !!)

there are   6+8 = 14 apples

2  +  14   + Orange = 3 (14)

  16 + O = 42     subtract 16 from both sides to find # O

a. The area of the graph is 18*pi.

b. b/a = -10.

One solution is (x,y) = (2 + sqrt(3), 2 - sqrt(3)).

[1, -2], [2, -4], [3, -6]

(a) 101^2 - 97^2 + 93^2 - 89^2 + ... + 5^2 - 1^2 = 101 + 97 + 93 + 89 + ... + 5 + 1 = 1326.

(b) (a + (2n + 1)d)^2 - (a + (2n)d)^2 + ... + (a + d)^2 - a^2 = (a + (2n + 1) d + (a + 2nd + ... + a + d) + a = n(3a + (n + 2)d).