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Let
\(f(x)=x+5\)
and let
\(g(x)=x^2+1\).
Let
\(p(x)=g(x)+f(x)\)
and let
\(q(x)=g(x)-f(x)\).
Find
\(p(x)\cdot q(x)\).

\(\begin{array}{|rcll|} \hline p(x)\cdot q(x) &=& \Big( g(x)+f(x) \Big) \Big( g(x)-f(x) \Big) \\ &=& [g(x)]^2-[f(x)]^2 \\ &=& (x^2+1)^2-(x+5)^2 \\ &=& x^4+2x^2+1 -(x^2+10x+25) \\ &=& x^4+2x^2+1 -x^2-10x-25 \\ &\mathbf{=}& \mathbf{x^4+x^2-10x-24} \\ \hline \end{array}\)

Using trigonometry to find length AB?

\(\begin{array}{|rcll|} \hline \cos(\alpha) &=& \dfrac{18}{70-x} \qquad (1)\\\\ \cos(\alpha) &=& \dfrac{18+24}{70} = \dfrac{42}{70} \qquad (2) \\\\ \cos(\alpha) =\dfrac{18}{70-x} &=& \dfrac{42}{70} \\\\ \dfrac{18}{70-x} &=& \dfrac{42}{70} \\\\ \dfrac{70-x}{18} &=& \dfrac{70}{42} \\\\ 70-x &=& \dfrac{70\cdot 18}{42} \\\\ x &=& 70- \dfrac{70\cdot 18}{42} \\\\ x &=& 70- 30 \\ \mathbf{x} & \mathbf{=}& \mathbf{40\ \text{cm}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 24^2+AB^2 &=& x^2 \quad & \quad x=40 \qquad \text{Pythagoras theorem} \\ 24^2+AB^2 &=& 40^2 \\ AB^2 &=& 40^2 - 24^2 \\ AB^2 &=& 1024 \quad & \quad 1024 = 32^2 \\ AB^2 &=& 32^2 \\ \mathbf{AB} & \mathbf{=}& \mathbf{32\ \text{cm}} \\ \hline \end{array}\)

Thank you so much! 

Because they are not acute enough to understand the joke?

2) See the answer here:  https://web2.0calc.com/questions/need-help-fast-please_1

1)

Solve the following system for x, y and z:
{x + y + 3 z = 10 | (equation 1)
-4 x + 2 y + 5 z = 7 | (equation 2)
k x + z = 3 | (equation 3)

Swap equation 1 with equation 2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
x + y + 3 z = 10 | (equation 2)
k x + 0 y+z = 3 | (equation 3)

Add 1/4 × (equation 1) to equation 2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+(3 y)/2 + (17 z)/4 = 47/4 | (equation 2)
k x + 0 y+z = 3 | (equation 3)

Multiply equation 2 by 4:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
k x + 0 y+z = 3 | (equation 3)

Add k/4 × (equation 1) to equation 3:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+(k y)/2 + ((5 k)/4 + 1) z = (7 k)/4 + 3 | (equation 3)

Multiply equation 3 by 4:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+2 k y + (5 k + 4) z = 7 k + 12 | (equation 3)

Subtract k/3 × (equation 2) from equation 3:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+0 y+(4 - (2 k)/3) z = 12 - (26 k)/3 | (equation 3)

Multiply equation 3 by 3/2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+0 y+(6 - k) z = 18 - 13 k | (equation 3)

Divide equation 3 by 6 - k:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y + 17 z = 47 | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Subtract 17 × (equation 3) from equation 2:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+6 y+0 z = (-6 (29 k - 4))/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Divide equation 2 by 6:
{-(4 x) + 2 y + 5 z = 7 | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(4 x) + 0 y+5 z = (65 k - 50)/(k - 6) | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Subtract 5 × (equation 3) from equation 1:
{-(4 x)+0 y+0 z = 40/(k - 6) | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

Divide equation 1 by -4:
{x+0 y+0 z = (-10)/(k - 6) | (equation 1)
0 x+y+0 z = (4 - 29 k)/(k - 6) | (equation 2)
0 x+0 y+z = (13 k - 18)/(k - 6) | (equation 3)

x = -10/(k - 6)
y = (4 - 29 k)/(k - 6)
z = (13 k - 18)/(k - 6)     k=6, for which there is NO solution.

See a couple of solutions here: https://web2.0calc.com/questions/a-bag-contains-some-marbles-each-of-which-is-one-of-four-colors-red-white-blue-and-green-at-least-one-of-the-marbles-is-red-the-comp

And here: https://artofproblemsolving.com/wiki/index.php?title=1996_AHSME_Problems/Problem_26

2.
The inverse of a modulo 39 is b.
What is the inverse of 4a modulo 39 in terms of b?
Give your answer as an expression in terms of b.

\(\begin{array}{|rcl|cl|} \hline 4a(4a)^{-1} &\equiv & 1 \pmod{39} \quad |\quad : (4a) \\\\ (4a)^{-1} &\equiv & \dfrac{1}{4a} \pmod{39} \\\\ (4a)^{-1} &\equiv & 4^{-1}a^{-1} \pmod{39} \quad | \quad a^{-1} = b \\\\ (4a)^{-1} &\equiv & 4^{-1} b \pmod{39} && 4^{-1} \pmod{39} \quad | \quad \gcd(4,39)=1 \\\\ && &\equiv & 4^{\phi(39)-1} \quad | \quad \phi(39)= 24 \\\\ && &\equiv & 4^{24-1} \pmod{39} \\\\ && &\equiv & 4^{23} \pmod{39} \\\\ && &\equiv & 10 \pmod{39} \\\\ \mathbf{(4a)^{-1} } & \mathbf{\equiv} & \mathbf{10 b \pmod{39} } \\ \hline \end{array}\)

3.

Let x and y be integers satisfying 41x+53y=12.
Find the residue of x modulo 53.
(Express your answer as an integer from 0 to 52.)

\(\small{ \begin{array}{|rcl|cl|} \hline 41x+53y &=& 12 \quad |\quad \cdot (-1) \\ -41x-53y &=& -12 \\ -41x &=& -12 +53y \\ \\ \hline \\ -41x &\equiv & -12 \pmod{53} \quad |\quad : (-41) \\\\ x &\equiv & \dfrac{-12}{-41} \pmod{53} \\\\ x &\equiv & \dfrac{12}{41} \pmod{53} \\\\ x &\equiv & 12\cdot 41^{-1} \pmod{53} && 41^{-1} \pmod{53} \quad | \quad \gcd(41,53)=1 \\\\ && &\equiv & 41^{\phi(53)-1} \quad | \quad \phi(53)= 52 \\\\ && &\equiv & 41^{52-1} \pmod{53} \\\\ && &\equiv & 41^{51} \pmod{53} \\\\ && &\equiv & 22 \pmod{53} \\\\ x &\equiv & 12\cdot 22 \pmod{53} \\\\ x &\equiv & 264 \pmod{53} \\\\ \mathbf{x} & \mathbf{\equiv} & \mathbf{52 \pmod{53}} \\ \hline \end{array} }\)

Proof:

\(\begin{array}{|rclr|} \hline 41x+53y &=& 12 \quad & | \quad \pmod{53} \\ 41x \pmod{53}+53y \pmod{53} &=& 12 \pmod{53} \quad & | \quad x \equiv 52 \pmod{53} \\ 41\cdot52 \pmod{53}+\underbrace{53y}_{\equiv 0\pmod{53} } \pmod{53} &=& 12 \pmod{53} \\ 41\cdot52 \pmod{53}+0 &=& 12 \pmod{53} \\ 2132 \pmod{53} &=& 12 \pmod{53} \\ 12 \pmod{53} &=& 12 \pmod{53}\ \checkmark\\ \hline \end{array} \)

Here's the question:

a)

\(\begin{array}{|lrcll|} \hline & y &=& a(x-h)^2 + k \quad & | \quad h = 0,\ k = 5 \\ & y &=& a(x-0)^2 + 5 \\ & \mathbf{y} & \mathbf{=}& \mathbf{ax^2 + 5} \\\\ P(5,4): & 4 &=& a5^2 + 5 \\ & 25a + 5 &=& 4 \\ & 25a &=& 4-5 \\ & 25a &=& -1 \quad & | \quad \\ & a &=& -\frac{1}{25} \\ & a &=& -0.04\\\\ \boxed{y=-0.04x^2+5} \\ \hline \end{array} \)

b)

The highway is on level y = 0

You can write the polynomial as:

f(x) = x(ax²+bx+c)     This form immediately gives f(0) = 0.

Now use the other conditions to find constants a, b and c:

15 = -1( (-1)²a + (-1)b +c)   or  15 = -a + b - c   ...(1)

-5 = 1(1²a + 1b + c)  or  -5 = a + b + c    ...(2)

12 = 2(2²a + 2b + c)  or  12 = 8a + 4b + 2c   or  6 = 4a + 2b + c...(3)

Add (1) and (2) to get:  10 = 2b  or  b = 5

Substitute this into (2) and (3) and rearrange:

-10 = a + c   ...(2b)

-4 = 4a + c  ...(3b)

Subtract (2b) from (3b)

6 = 3a   or  a = 2

Substitute this into (2b)

-10 = 2 + c  or  c = -12

Now we have:  f(x) = x(2x² + 5x - 12) 

This factors as f(x) = x(2x - 3)(x + 4)

Hence the roots are: 0, 3/2, -4

You should check this by substituting the values -1, 0, 1 and 2 for x into the function to see they give the values specified in the question.

The polynomial $f(x)$ has degree 3. If
$f(-1) = 15$,
$f(0)= 0$,
$f(1) = -5$, and
$f(2) = 12$,
then what are the $x$-intercepts of the graph of $f$?

\(\begin{array}{|llcll|} \hline \boxed{f(x)=ax^3+bx^2+cx+d }\\\\ f(0) = 0: & 0 = a\cdot0^3+b\cdot 0^2+c\cdot 0 + d \\ & \mathbf{\boxed{d=0}} \\\\ \boxed{f(x)=ax^3+bx^2+cx+0 \\ f(x) = x(ax^2+bx+c) }\\\\ f(-1) = 15: & 15 = (-1)\left(a(-1)^2+b(-1)+c \right) \\ & 15 = (-1) (a-b+c ) \\ &\mathbf{ 15 = -a+b-c \qquad (1) } \\ f(1) = -5: & -5 = 1\cdot\left(a(1)^2+b(1)+c \right) \\ &\mathbf{ -5 = a+b+c \qquad (2) }\\\\ (1)+(2): & 15-5 =b+b \\ & 10 = 2b \\ & \mathbf{\boxed{b=5}} \\\\ \text{see }(1): & 15 = -a+b-c \quad b=5 \\ & 15 = -a+5-c \\ & 10 = -a-c \\ & \mathbf{c = -10-a \qquad (3)} \\\\ \boxed{f(x) = x(ax^2+bx+c) \\ f(x)=x(ax^2+5x-10-a)} \\\\ f(2) = 12: & 12 = 2\cdot(a\cdot2^2+5\cdot 2-10-a ) \\ & 6 = 4a-a \\ & 6 = 3a \\ & \mathbf{\boxed{a=2}} \\\\ \text{see }(3): & c = -10-a \quad a=2 \\ & c = -10-2 \\ & \mathbf{\boxed{c=-12}} \\ \hline \end{array}\)

\(\text{The polynomial $f(x)$ of degree $3$ is:} \\ \boxed{f(x) = 2x^3 + 5x^2-12x \\ f(x) = x(2x^2+5x-12) }\)

\(\text{x-intercepts $f(x) = 0\ ?$}\\ \begin{array}{|rcll|} \hline \boxed{ f(x) = x(2x^2+5x-12) }\\\\ 0 &=& x(2x^2+5x-12) \\ \mathbf{x_1} &\mathbf{=}&\mathbf{ 0} \\\\ 2x^2+5x-12 &=& 0 \\ x &=& \dfrac{-5\pm \sqrt{25-4\cdot 2\cdot (-12) } }{2\cdot 2} \\ x &=& \dfrac{-5\pm \sqrt{25+96 } }{4} \\ x &=& \dfrac{-5\pm \sqrt{121} }{4} \\ x &=& \dfrac{-5\pm 11 }{4} \\\\ x_2 &=& \dfrac{-5+ 11 }{4} \\ x_2 &=& \dfrac{6}{4} \\ \mathbf{x_2} & \mathbf{=}& \mathbf{ 1.5 } \\\\ x_3 &=& \dfrac{-5- 11 }{4} \\ x_3 &=& -\dfrac{16}{4} \\ \mathbf{x_3} & \mathbf{=}& \mathbf{ -4 } \\ \hline \end{array} \)

The x-intrcepts are: \( x=-4,\ x=0,\ x=1.5\)

The graph:

If f(x)=x^5-1/3, find f^-1(-31/96).

\(\begin{array}{|rcll|} \hline \boxed{f\left(f^{-1}\left(\dfrac{-31}{96}\right)\right) = \dfrac{-31}{96} \qquad (1) } \\\\ f(x) &=& x^5-\dfrac{1}{3} \quad | \quad x = f^{-1}\left(\dfrac{-31}{96}\right) \\\\ f\Big(f^{-1}\left(\dfrac{-31}{96}\right)\Big) &=& \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5-\dfrac{1}{3} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5-\dfrac{1}{3} = \dfrac{-31}{96} \quad | \quad \text{ see } (1) \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{1}{3} - \dfrac{31}{96} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{96-3\cdot 31}{3\cdot 96} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{3}{3\cdot 96} \\\\ && \left(f^{-1}\left(\dfrac{-31}{96}\right) \right)^5 = \dfrac{1}{96} \\\\ && f^{-1}\left(\dfrac{-31}{96}\right) = \sqrt[5]{\dfrac{1}{96}} \\\\ && f^{-1}\left(\dfrac{-31}{96}\right) = \sqrt[5]{0.01041666667} \\\\ && \mathbf{ f^{-1}\left(\dfrac{-31}{96}\right) = 0.40137078088 } \\ \hline \end{array}\)

As far as I can see, everything looks OK, except for 4-year depreciations in c and d. You have them switched around. The 4-year depreciation using the exponential model would be: 0.7^4 x $54,000 =$12,965 in "c". The 4-year depreciation using the linear model would be: 54,000 - (4 x $6,750) =$27,000 in "d". And that is it. 

x =[100 - 20] / 2 =40 degrees - or "b" on your list.

360 - 85 -115 - 50 =110 degrees - Length of arc opposite angle "5"

The measure of angle "5" =1/2[110 + 115] =112.5 degrees - or "c" on your list.