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1. p(x)*q(x) = x^3 + 6x^2 + x + 5.

2. c = -5.

The angles are 59, 60, 61, 62, 63.

The volume of the frustum is 100/3*pi.

(b) x^3 + 4*x^2*y - 2y^3 = 0.

(c) r^3 - 4r - 2= 0.

(d) cos 36 = (-2 + sqrt(2))/2, cos 72 = (-2 - sqrt(2))/2.

(b) x^3 - 3*x^2*y + y^3 = 0.

(c) r^3 - 3r + 1 = 0.

(d) cos 36 = (-1 + sqrt(3))/2, cos 72 = (-1 - sqrt(3))/2.

This question went through months ago.

Trawl back through the answers pages and you might be able to find the solution.

5^2=5²=25

1.5a=b -----> I

b=1/a------->II

Substitute for b

1.5a=1/a

1.5a-(1/a)=0

Multiply both sides by (a)

(1.5a²)-1=0

1.5a²=1

a²=1/1.5

a= sqrt(1/1.5)

a=sqrt(2/3)

a=0.816497

---------------------

ab=1

b=1/a

b=1/(sqrt(2/3))

b=1.224745

How many different integers can be expressed as the sum of three distinct members of the set {1,4,7,10,13,16,19}?

 \(\begin{array}{|r|r|r|} \hline 1.)& ( 1, 4, 7) & sum = 12 \\ 2.)& ( 1, 4, 10) & sum = 15 \\ 3.)& ( 1, 4, 13) & sum = 18 \\ 4.)& ( 1, 4, 16) & sum = 21 \\ 5.)& ( 1, 4, 19) & sum = 24 \\ 6.)& ( 1, 7, 19) & sum = 27 \\ 7.)& ( 1, 10, 19) & sum = 30 \\ 8.)& ( 1, 13, 19) & sum = 33 \\ 9.)& ( 1, 16, 19) & sum = 36 \\ 10.)& ( 4, 16, 19) & sum = 39 \\ 11.)& ( 7, 16, 19) & sum = 42 \\ 12.)& (10, 16, 19) & sum = 45 \\ 13.)& (13, 16, 19) & sum = 48 \\ \hline \end{array} \)

The question specifies that the real number x satisfies  x^2 - 5x + 6<0  (not x^2 - 5x - 6 < 0). 

This means x is in the range from 2 to 3.  Substitute these values into x^2 + 5x + 6 to find the end points of the required set of values. 

G is the centroid of ABC and D and E are the midpoints of AB and AC respectively. AG and DE intersect at M. Find GM/GA

\(\dfrac{GM}{GA} = \dfrac{1}{4} \)

My answer is 1.8 but im not sure about the direction. 

Your question is incomplete.  

Do you know what a convex quadrilateral is?  Perhaps you would like to share this knowledge with other forum users.?

Have you tried playing with a rough sketch. If so you will be able to discound some of the possibilies straight away.

Which ones can you discount straight away, and why?

This question has been asked many times in the recent past.

The earlier answers are easy to find  Sinclairdragon428  posted them.  (He likely posted this time too)

Why do want to change your name? Did you have transgender surgery?

Find the smaller angle formed between the hour hand and the minute hand at time 6:48 pm.

\(\begin{array}{rcll} \Delta\alpha^{\circ} &=& 330 \times t^h \pmod{360^{\circ}} \quad | \quad \mathbf{t^h=6+\dfrac{48}{60} = 6.8\ h } \\ &=& 330 \times 6.8 \pmod{360^{\circ}} \\ &=& 2244^{\circ} \pmod{360^{\circ}} \\ &=& \mathbf{84^{\circ}} \\ \end{array} \)

How many weeks are there in \(8!\) minutes?

\(\begin{array}{|rcll|} \hline 8!\ \text{min.} &=& 1*2*3*4*5*6*7*8\ \text{min.} \\ &=& 40320\ \text{min.} \\ \hline && 40320\ \text{min.} \times \dfrac{1\ \text{h}}{60\ \text{min.}} \times \dfrac{1\ \text{d}}{24\ \text{h}} \times \dfrac{1\ \text{week}}{7\ \text{d}} \\ &=& \dfrac{ 40320} {60\cdot 24\cdot 7} \ \text{weeks} \\ &=& \mathbf{4 \ \text{weeks} } \\ \hline \end{array} \)

8! minutes are 4 weeks.

The sum of the coordinates is 4 + 11 = 15.

The area of PCQA is 15.

The real soluitons are x = -4, 0, and 4.

1. The probability is 11/2188.

2. The probability is 5/21.

3. The number of ways is 38226.