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We can use the sine law to find P1 P2

By the sine law, P1 P2 = sin 15.  We can also find P1 P_3 = sin 60, P1 P4 = sin 90, etc.

So (P1 P2)^2 + (P1 P3)^2 + … + (P11 P12)^2 = 12 (sin 15)^2 + 12 (sin 30)^2 + 12 (sin 45)^2 + … + 12 (sin 90)^2 = 42.

\(\frac{2}{3}\)

Important: When a tangent and a secant, two secants, or two tangents intersect outside a circle then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

360 - ( 130 + 75 ) = 155

x = ( 155 - 75 ) / 2 

I need help with this problem.

Show all work to factor x^4 − 17x^2 + 16 completely.

Hello Guest!

\(x^4-17x^2 +16=0\)

We determine the zeros.

Exchange:  \(x^2=u\)

\(u^2-17u+16=0\\ u^2-m\cdot u+n =0\)

\(u=-\frac{m}{2}\pm \sqrt{(\frac{m}{2})^2-n}\\ u=8.5\pm \sqrt{8.5^2-16}\\ u=8.5\pm7.5\)

\(u_1=16\\ u_2=1\)

Reinstate:  \(x^2=u\\ x=\sqrt{u}\)

\(x_1=+\sqrt{16}\\ x_2=-\sqrt{16}\\ x_3=+\sqrt{1}\\ x_4=-\sqrt{1}\)

\(x_1=4\\ x_2=-4\\ x_3=1\\ x_4=-1\)

\(x^4-17x^2 +16=(x-4)(x+4)(x-1)(x+1)\)

  !

I need help matching each expression.

Factor the following:
x^4-17 x^2+16

x^4-17 x^2+16==(x^2)^2-17 x^2+16:
(x^2)^2-17 x^2+16

The factors of 16 that sum to -17 are -1 and -16. So, (x^2)^2-17 x^2+16==(x^2-1) (x^2-16):
(x^2-1) (x^2-16)

x^2-16==x^2-4^2:
(x^2-1) (x^2-4^2)

Factor the difference of two squares. x^2-4^2==(x-4) (x+4):
(x-4) (x+4) (x^2-1)

x^2-1==x^2-1^2:
(x^2-1^2) (x-4) (x+4)

Factor the difference of two squares. x^2-1^2==(x-1) (x+1):

(x-1) (x+1) (x-4) (x+4)

How do I do a proper link?

Thanks

To begin this, let's find the slope of that line. The slope is: 15/4. Now, we need to find the y intercept. To do this, we substitute (0,y) for (-8, -6). Solve for the slope (and set it equal to 15/4), and we get 24. Therefore, the equation of the line is y=15/4x+24.

Here is the graph:

I suspect this question should read something like:

If sinθ=3/5, find θ, where 0º < θ < 360º  (or, perhaps, some other upper limit on θ).

I'm have trouble understanding: "where 0". Could you please clarify? Thanks!

Thank you, guest :)

\(\displaystyle \text{If}\;\;x^{2}+px+q=0 \;\;\text{ has real (distinct) roots, then }\;\; p^{2}-4q>0.\)

\(\displaystyle x^{2}+px+q+(x+a)(2x+p)\\=x^{2}+px+q+2x^{2}+x(2a+p)+ap\\ =3x^{2}+x(2a+2p)+q+ap=0\)

\(\displaystyle \text{For this to have real (distinct) roots, it's necessary that}\\ (2a+2p)^{2}-12(q+ap)>0, \\ \text{and since}\\ = (2a+2p)^{2}-12(q+ap)\\=4a^{2}+4p^{2}+8ap-12q-12ap \\= 4a^{2}+p^{2}-4ap+3p^{2}-12q\\=(2a-p)^{2}+3(p^{2}-4q)\\ \text{this will be the case for any real number a.} \)

In the diagram, \(\angle U = 30^\circ\), arc XY is \(170^\circ\), and arc VW is \(110^\circ\).

Find arc WY, in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\text{ In $\triangle $ UWY}} \\ \hline \mathbf{180^\circ} &=& \mathbf{U + \left(90^\circ-\dfrac{170^\circ}{2}+ 90^\circ-\dfrac{WY}{2}\right)} \\ && \quad \mathbf{ +\left(90^\circ-\dfrac{110^\circ}{2}+ 90^\circ-\dfrac{WY}{2} \right)} \\\\ 180^\circ &=& U + \left(180^\circ-85^\circ-\dfrac{WY}{2}\right) \\ && \quad + \left(180^\circ-55^\circ-\dfrac{WY}{2} \right) \\\\ 180^\circ &=& U + 95^\circ-\dfrac{WY}{2} + 125^\circ-\dfrac{WY}{2} \\\\ 180^\circ &=& U + 220^\circ - WY \quad | \quad U=30^\circ \\ 180^\circ &=& 30^\circ + 220^\circ - WY \\ 180^\circ &=& 250^\circ - WY \\ WY &=& 250^\circ - 180^\circ \\ \mathbf{WY} &=& \mathbf{70^\circ} \\ \hline \end{array}\)

The deck for a card game contains 30 cards. 10 are red, 10 yellow, 5 blue, and 1 green, and 4 are wild cards. Each player is randomly dealt a five-card hand.

a) What is the probability that a hand will contain exactly two wild cards?

\(\text{Prob of exactly 2 wild cards}\\ =\frac{\text{number with exactly 2 wild cards}}{\text{number of possible hands}}\\ =\frac{4C2 * 26C3}{30C5}\\ =\frac{4!26!}{2!2!3!23!}\times \frac{5!25!}{30!}\\ =\frac{24*25*26}{1}\times \frac{2*3*4*5}{26*27*28*29*30}\\ =\frac{6*5}{1}\times \frac{2*3*4*5}{27*7*29*6}\\ =\frac{5}{1}\times \frac{2*4*5}{9*7*29}\\ = \frac{200}{1827}\\ \)

I have not checked my working.

a is good

but

b and c are not right. B is almost right   

Not sure what you mean but arcsine is just the inverse of sin

\(asin[sin\theta]=\theta \)

sin (angle)=ratio       (number)

asin (ratio) = angle

sometimes   asin     is  written as   \(sin^{-1}\)

arc sine  is the same as inverse sine

On a calculator you would normally press Inv or 2nd   and then the sine button if you want inverse sin.

We have Ax + By = 3   (1)

If this passes through (7, -2) then  7A - 2B = 3   (2)

We can rearrange equation (1) in the form  y = -(A/B)x + 3/B  The slope of this equation is -A/B

The equation x + 3y = -5 can be rearranged as  y = -(1/3)x -5/3  the slope of which is -1/3

For the two lines to be parallel the slopes must be the same, hence  -A/B = -1/3   (3)

Now you have two equations ((2) and (3)) that can be solved for A and B.

Can you take it from here?

Rearrange the equations to get y = ...

-2x + y = k  becomes  y = k + 2x                 (1)

0.5x + y = 14 becomes  y = 14 - 0.5x          (2)

Now equate (1) and (2)

k + 2x = 14 - 0.5x   or  k = 14 - 2.5x  (3)

Replace x in (3) by -8.4 to find k.

Let
\(A = \dfrac{10^{1998} + 1}{10^{1999} + 1}\) and \(B=\dfrac{10^{1999} + 1}{10^{2000} + 1}\)

Which value is larger?

\(\begin{array}{|rcll|} \hline \mathbf{B} &=& \mathbf{ \dfrac{10^{1999} + 1}{10^{2000} + 1} } \\\\ B &=& \dfrac{10*10^{1998} + 1}{10*10^{1999} + 1} \\\\ B &=& \dfrac{10*\left(10^{1998} + \frac{1}{10} \right)}{10*\left(10^{1999} + \frac{1}{10}\right)} \\\\ \mathbf{B} &=& \mathbf{ \dfrac{10^{1998} + \frac{1}{10} }{10^{1999} + \frac{1}{10}} } \\ \hline \end{array}\)

\(\Rightarrow A \gt B\)

6C2*4 + 4C2*6

2.
Let \(f(x)\)  be a polynomial such that \(f(0) = 4\), \(f(1) = 5\), and \(f(2)=10\).
Find the remainder \((=r)\) when \(f(x)\) is divided by \(x(x - 1)(x - 2)\).

\(\begin{array}{|lrclrcl|} \hline & f(x) &=& q(x)*x(x - 1)(x - 2) + r \quad | \quad \mathbf{r=ax^2+bx+c } \\ & \mathbf{ f(x) } &=& \mathbf{ q(x)*x(x - 1)(x - 2) + ax^2+bx+c } \\ \hline x=0:& \mathbf{ f(0) } &=& \mathbf{ q(0)*0(0 - 1)(0 - 2) + a*0^2+b*0+c } \\ & f(0)&=& q(0)*0 +c \\ & f(0)&=& c \quad | \quad f(0) = 4 \\ & 4&=& c \\ & \mathbf{c} &=& \mathbf{4} \\ \hline x=1:& \mathbf{ f(1) } &=& \mathbf{ q(1)*1(1 - 1)(1 - 2) + a*1^2+b*1+c } \\ & f(1)&=& q(1)*0 a+b+4 \\ & f(1)&=& a+b+4 \quad | \quad f(1) = 5 \\ & 5&=& a+b+4 \\ & \mathbf{a+b} &=& \mathbf{1} \qquad (1) \\ \hline x=2:& \mathbf{ f(2) } &=& \mathbf{ q(2)*2(2 - 1)(2 - 2) + a*2^2+b*2+c } \\ & f(2)&=& q(2)*0 +4a+2b+4 \\ & f(2)&=& 4a+2b+4 \quad | \quad f(2) = 10 \\ & 10&=& 4a+2b+4 \quad | \quad :2 \\ & 5&=& 2a+b+2 \\ & \mathbf{2a+b} &=& \mathbf{3} \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{a+b} &=& \mathbf{1} \qquad (1) \\ & \mathbf{2a+b} &=& 3 \qquad (2) \\ \hline (2)-(1): & 2a+b -(a+b) &=& 3-1 \\ & \mathbf{a} &=& \mathbf{2} \\ \hline & b &=& 1-a \\ & b &=& 1-2 \\ & \mathbf{b} &=& \mathbf{-1} \\ \hline & r &=& ax^2+bx+c \\ & \mathbf{ r } &=& \mathbf{2x^2-x+4} \\ \hline \end{array} \)

The remainder when \(f(x)\) is divided by \(x(x - 1)(x - 2)\) is  \(\mathbf{2x^2-x+4}\)

1.

\(f(x) = x^{10}+5x^9-8x^8+7x^7-x^6-12x^5+4x^4-8x^3+12x^2-5x-5\)
 

Without using long division, find the remainder \((=r)\)  when \(f(x)\)  is divided by \(x^2-1\).

Let \(x^2-1 = (x-1)(x+1)\)

\(\begin{array}{|lrclrcl|} \hline & f(x) &=& q(x)(x-1)(x+1) + r \quad | \quad \mathbf{r=ax+b} \\ & \mathbf{ f(x) } &=& \mathbf{ q(x)(x-1)(x+1) + ax+b } \\ \hline x=1:& \mathbf{ f(1) } &=& \mathbf{ q(1)(1-1)(1+1) + a*1+b } \\ & f(1)&=& q(1)*0 + a+b \\ & f(1)&=& a+b \\ &&& \small {f(1) = 1*1^{10}+5*1^9-8*1^8+7*1^7-1^6-12*1^5+4*1^4-8*1^3+12*1^2-5*1-5 }\\ &&& f(1) = 1+5-8+7-1-12+4-8+12-5-5 \\ &&& f(1) = -8+7+4-8-5 \\ &&& f(1) = 11-21 \\ &&& f(1) = -10 \\ & -10 &=& a+b \\ & \mathbf{a+b} &=& \mathbf{-10} \qquad (1) \\ \hline x=-1:& \mathbf{ f(-1) } &=& \mathbf{ q(-1)(-1-1)(-1+1) + a*(-1)+b } \\ & f(-1)&=& q(-1)*0 - a+b \\ & f(-1)&=& - a+b \\ &&& \small{f(-1) = 1*(-1)^{10}+5*(-1)^9-8*(-1)^8+7*(-1)^7-(-1)^6-12*(-1)^5+4*(-1)^4-8*(-1)^3+12*(-1)^2-5*(-1)-5} \\ &&& f(-1) = 1-5-8-7-1+12+4+8+12+5-5 \\ &&& f(-1) =12+4 \\ &&& f(-1) = 16 \\ & 16 &=& - a+b \\ & \mathbf{-a+b} &=& \mathbf{16} \qquad (2)\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{a+b} &=& \mathbf{-10} \qquad (1) \\ & \mathbf{-a+b} &=& \mathbf{16} \qquad (2)\\ \hline (1)+(2): & a+b-a+b&=&-10+16 \\ & 2b&=& 6 \\ & \mathbf{b} &=& \mathbf{3} \\ \hline (1)-(2): & a+b-(-a+b) &=& -10-16 \\ & 2a &=& -26 \\ & \mathbf{a} &=& \mathbf{-13} \\ \hline & r &=& ax+b \\ & \mathbf{ r } &=& \mathbf{-13x+3} \\ \hline \end{array}\)

The remainder when \(f(x)\) is divided by \(x^2-1\) is \(\mathbf{-13x+3}\)