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The graphs of $y=3-x^2+x^3$ and $y=1+x^2+x^3$ intersect in multiple points. Find the maximum difference between the $y$-coordinates of these intersection points.

\(\begin{array}{|rcll|} \hline 3-x^2+\not{x^3} &=& 1+x^2+ \not{x^3} \\ 3-x^2 &=& 1+x^2 \quad & | \quad -1-x^2 \\ 3-x^2 -1-x^2 &=& 0 \\ 2-2x^2 &=& 0 \\ 2(1-x^2) &=& 0 \quad & | \quad : 2 \\ 1-x^2 &=& 0 \\ x^2 &=& 1 \quad & | \quad \sqrt() \\ x &=& \pm\sqrt{1} \\\\ x_1 &=& 1 \\ y_1 &=& y(1) \\ &=& 1+1+1 \\ \mathbf{y_1 } &\mathbf{=}& \mathbf{3} \\\\ x_2 &=& -1 \\ y_2 &=& y(-1) \\ &=& 1-1+1 \\ \mathbf{y_2} &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline y_1 - y_2 &=& 3-1 \\ &=& 2 \\ \hline \end{array}\)

The maximum difference between the  y-coordinates of these intersection points is 2

This graph should help:

What is the value of x?

sin32°=cosx

Enter your answer in the box

sin(32) = cos(x)

x = 90 - 32

x = 58, so that:

sin(32) = cos(58)

help me, please.

Exercises: Solve the exponential Equation rounding decimal answers to the thousandths place.

3^x=12

\(\begin{array}{|rcll|} \hline 3^x &=& 12 \quad & | \quad \log() \text{ both sides} \\ \log{(3^x)} &=& \log(12) \\ x\log{(3)} &=& \log(12) \quad & | \quad : \log{(3)}\\\\ x &=& \dfrac{\log(12)} {\log{(3)}} \\\\ x &=& \dfrac{1.07918124605} {0.47712125472} \\\\ x &=& 2.26185950714 \\\\ \mathbf{x} &\mathbf{\approx}& \mathbf{2.262} \\ \hline \end{array}\)

3^x=12      Take the log of both sides

x = Log(12) / Log(3)

x = 2.262

Suppose that p is prime and

\(1007_p+306_p+113_p+125_p+6_p=142_p+271_p+360_p. \)

1007_p+306_p+113_p+125_p+6_p=142_p+271_p+360_p.

How many possible values of  p are there?

\(\small{ \begin{array}{|rcll|} \hline 1007_p+306_p+113_p+125_p+6_p &=& 142_p+271_p+360_p \\\\ \underbrace{p^3+7}_{=1007_p}+\underbrace{3p^2+6}_{=306_p}+ \underbrace{p^2+p+3}_{=113_p} + \underbrace{p^2+2p+5}_{=125_p} +\underbrace{6}_{=6_p} &=& \underbrace{p^2+4p+2}_{=142_p}+ \underbrace{2p^2+7p+1}_{=271_p} + \underbrace{3p^2+6p}_{=360_p} \\ p^3+ 3p^2+ p^2+ p^2 +p +2p+7 +6 +3 +5 + 6 &=& p^2+ 2p^2+ 3p^2 +4p+ 7p+6p +2 + 1 \\ p^3+ 5p^2+3p +27 &=& 6p^2 +17p +3 \\ p^3+ 5p^2-6p^2+3p-17p +27-3 &=& 0 \\ p^3-p^2-14p +24 &=& 0 \\ (p-2)(p^2+p-12) &=& 0 \\ (p-2)(p-3)(p+4) &=& 0 \\ \hline \end{array} }\)

The only primes that will work are \( p=2 \) or \(p=3\)

Medians are drawn from point A and point B in this right triangle to divide segments BC and AC in half, respectively. The lengths of the medians are 6 and \( 2\sqrt{11}\) units, respectively.

How many units are in the length of segment AB.

\(\text{Let $c = AB $} \\ \text{Let $a = BC $} \\ \text{Let $b = CA $} \)

\(\begin{array}{|lrcll|} \hline (1) & \left(\dfrac{1}{2}a \right)^2 + b^2 &=& 6^2 \\ (2) & a^2 + \left(\dfrac{1}{2}b \right)^2 &=& (2\sqrt{11})^2 \\ \hline (1) + (2): & \left(\dfrac{1}{2}a \right)^2 + b^2 + a^2 + \left(\dfrac{1}{2}b \right)^2 &=& 6^2 + (2\sqrt{11})^2 \\ & \dfrac{5}{4}a^2 + \dfrac{5}{4}b^2 &=& 36+44 \\ & \dfrac{5}{4}(a^2 + b^2) &=& 80 \quad & | \quad a^2+b^2=c^2 \\ & \dfrac{5}{4}c^2 &=& 80 \\ & c^2 &=& \dfrac{4\cdot 80}{5} \\ & c^2 &=& 4\cdot 16 \\ & c &=& 2\cdot 4 \\ & \mathbf{c} & \mathbf{=} & \mathbf{8} \\ \hline \end{array}\)

The length of segment AB = 8

Points $P$, $Q$, and $R$ are on sides $\overline{AB}$, $\overline{CD}$, and $\overline{BC}$ of rectangle $ABCD$ as shown below such that $BP = 4$, $CQ = 18$, $\angle PRQ = 90^\circ$, and $CR = 2BR$.

Find $PQ$. 

\(\text{Let $x = \overline{PQ}$} \\ \text{Let $y = \overline{BC}=\overline{AD}$} \\ \text{Let $u = \overline{RQ}$} \\ \text{Let $v = \overline{RP}$} \)

\(\begin{array}{|lrcll|} \hline & x^2 &=& y^2+(18-4)^2 \\ & x^2 &=& y^2+(14)^2 \\ & x^2 &=& y^2+196\\ (1) & y^2 &=& x^2-196 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline v^2 &=& 4^2+ \left( \dfrac{1}{3}y \right)^2 \\ &=& 16+ \dfrac{1}{9}y^2 \\\\ u^2 &=& 18^2 + \left( \dfrac{2}{3}y \right)^2 \\ &=& 324 + \dfrac{4}{9}y^2 \\\\ x^2 &=& u^2+v^2 \\ &=& 324 + \dfrac{4}{9}y^2 + 16+ \dfrac{1}{9}y^2 \\ &=& 340 + \dfrac{5}{9}y^2 \\ x^2 &=& 340 + \dfrac{5}{9}y^2 \quad &| \quad y^2 = x^2-196 \\ x^2 &=& 340 + \dfrac{5}{9}(x^2-196) \\ x^2 &=& 340 + \dfrac{5}{9}x^2- \dfrac{5\cdot 196}{9} \\ x^2+ \dfrac{5}{9}x^2 &=& 340 - \dfrac{5\cdot 196}{9} \\ \dfrac{4}{9}x^2 &=& 340 - \dfrac{5\cdot 196}{9} \quad &| \quad \cdot \dfrac{9}{4} \\ x^2 &=& \dfrac{9\cdot 340}{4} - \dfrac{5\cdot 196}{4} \\ x^2 &=& \dfrac{9\cdot 340-5\cdot 196}{4} \\ x^2 &=& 520 \\ x^2 &=& 4\cdot 130 \\ x &=& 2\cdot\sqrt{ 130} \\ \mathbf{x} & \mathbf{\approx}& \mathbf{22.80} \\ \hline \end{array}\)

cos(A - 90°) = cos(A)cos(90°) + sin(A)sin(90°)  or  cos(A - 90°) = cos(A)*0 + sin(A)*1 or cos(A - 90°) = sin(A)

If A = x + 34, then sin(x + 34) can be written as cos(x + 34 - 90) or cos(x - 56)

So if sin(x+34) = cos(2x + 20) we could also write  cos(x - 56) = cos(2x + 20) which means that x - 56 = 2x + 20 

rearranging, we have: x = -76°  

Check:

sin(-76+34) = -0.6691

cos(2*(-76)+20) = -0.6691

If you want a value of x between 0 and 360, simply add 360 to x (i.e 360 - 76 = 284).

sin R will be the same as cos P, and cos R will be the same as sin P.  This should be enough information for you to find sin R cos R.

There are 6 solutions when y = -3 and when y = -4.  i.e. a horizontal line at y = -3 cuts the curve in 6 places (similarly for y = -4).  Hence c = -3 and c = -4. I'll leave you to add them together!

The expression can be written as \(\frac{(x+1)^2}{(x-1)(x+3)}\)    which is undefined when the denominator is zero.

This occurs when x = 1 and x = -3, so the product is -3.

.

Tan(x) = 5 / 20

x =arctan(5/20)

x = ~14.04 degrees.

Simplify the following:
(12 i + 5)/(-3 i + 2)

Multiply numerator and denominator of (12 i + 5)/(-3 i + 2) by 2 + 3 i:
((12 i + 5) (3 i + 2))/((-3 i + 2) (3 i + 2))

(2 - 3 i) (2 + 3 i) = 2×2 + 2×3 i - 3 i×2 - 3 i×3 i = 4 + 6 i - 6 i + 9 = 13:
((12 i + 5) (3 i + 2))/13

(5 + 12 i) (2 + 3 i) = 5×2 + 5×3 i + 12 i×2 + 12 i×3 i = 10 + 15 i + 24 i - 36 = -26 + 39 i:
(39 i - 26)/13

Factor 13 out of 39 i - 26 giving 13 (3 i - 2):
(13 (3 i - 2))/13

(13 (3 i - 2))/13 = 13/13×(3 i - 2) = 3 i - 2:

 =3i - 2    OR    -2 + 3i

The maximum side must be  < 20/2 = 10, because the sum of the 2 shorter sides must be > than the longest side.

So, I would say 9 would the maximum side. The other 2 sides could be:

8 and 3

7 and 4

6 and 5 in that they all must be whole integers and different from each other.

50 - [26% x 50] =

50 - [0.26 x 50] =

50 -  13 = 37 points left.

7 / [7+8] x 90 = 42 degrees - the smaller angle

8 / [7+8] x 90 = 48 degrees - the larger angle.

10 / [10+3+7] x 180 = 90 degrees - the largest angle

7 / [10+3+7] x 180 = 63 degrees - the next largest angle

3 / [10+3+7] x 180 = 27 degrees - the smallest angle.

Okay yes that makes sense now thanks for the explanation and showing each step to solve.

Well, you make things pretty simple when no interest rates are involved!!.

40 years x 12 months= 480 months

$750 x 480 =$360,000 - this is how much you will have saved in 40 years or 480 months.

$360,000 / $2,500 = 144 - months that you will be withdrawing monthly payments of $2,500 each.

144 / 12 = 12 years - that your savings of $360,000 will run out.

65 + 12 = 77 years - your age when you run out of your retirement savings.

Thanks so much, CPhill!!!!!

Its the average before the extra credit.

Given that the point  (8, 8)  is on the graph of  \(y=\frac 14f\left(\frac 12x\right)\) , there is one point that must be on the graph of  y = f(x) . What is the sum of the coordinates of that point?

\(8=\frac 14f\left(\frac 12(8)\right)\\~\\ 8=\frac 14f\left(4\right)\\~\\ 4\cdot8=f\left(4\right)\\~\\ 32=f\left(4\right)\\~\\ f\left(4\right)=32\)

So  (4, 32)  must be a point on the graph of  y = f(x) .

4 + 32  =  36

Let  QR  be the base of the triangle.

Let's draw a height from  QR  to  P .

Now we have a right triangle.

sin( angle )  =  opposite / hypotenuse

sin 20°  =  h / 9

                            Multiply both sides of the equation by  9 .

9 sin 20°  =  h

h  =  9 sin 20°

Now we know the base and the height, so we can find the area.

area  =  (1/2)(base)(height)

area  =  (1/2)(14)(9 sin 20°)   square cm

area  =  63 sin 20°   square cm

area  ≈  21.5   square cm

The square route of 666 is 25.8069758011278803. I just calculated it.