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Let M and N be constants. The graphs of the lines x+5y=7 and 15x+MY=N are perpendicular and intersect at (-8,3) Enter the ordered pair (M,N)

The least common multiple will be 3*3*5*x

The highest common factor is  some factor of 45

so

\(\frac{3*3*5* \boxed{x} }{\text{some factor of 45}}=\frac{33}{1}\\ \frac{3*3*5*\boxed{11}}{\boxed{3*5}}=\frac{3*11}{1}\)

So the least common multiple is 3*3*5*11

The highest common factor is  3*5

so they are both multiples of 15 and ours must be a multiple of 11 as well ... 3*5*11=165

So I think the two numbers are  45 and 165

"A box contains 81 coins, only dimes and nickels. The amount of money in the box is $5.30. How many dimes and how many nickels are in the box?"

d + n = 81   (1)

10d + 5n = 530 divide by 5 to get 2d + n = 106   (2)

Subtract (1) from (2):   d = 106 - 81 = 25   (3)

Use (3) in (1):  25 + n = 81 or n = 81 - 25 = 56

There are 25 dimes and 56 nickels.

Does this help?

Two adjacent sides of a rhombus form a 60-degree angle. If each side of the rhombus measures 2 cm, what is the area of the rhombus, in square centimeters? Express your answer in simplest radical form.

In the diagram, BD = 5, CE = 4, [ABC] = 24, and [ADE] = 18.

Find [ACD].

\(\begin{array}{|lrcll|} \hline (1) & 24 + [ACD] &=& \dfrac{5h}{2} \\\\ (2) & 18 + [ACD] &=& \dfrac{4h}{2} \\ \hline (1)-(2): & 24 + [ACD] -(18 + [ACD]) &=& \dfrac{5h}{2}-\dfrac{4h}{2} \\\\ & 24 + [ACD] -18 - [ACD] &=& \dfrac{1h}{2} \\\\ & 24 -18 &=& \dfrac{h}{2} \\\\ & 6 &=& \dfrac{h}{2} \\\\ & \mathbf{h} &=& \mathbf{12} \\ \hline & 18 + [ACD] &=& \dfrac{4h}{2} \\ & 18 + [ACD] &=& 2h \quad & | \quad h=12 \\ & 18 + [ACD] &=& 2h \\ & 18 + [ACD] &=& 2*12 \\ & 18 + [ACD] &=& 24 \\ & [ACD] &=& 24-18 \\ & \mathbf{[ACD]} &=& \mathbf{6} \\ \hline \end{array}\)

Let  r1, r2, r3 and r4 be the roots of x^4 - 2x^3 - 5x^2 + 4x - 1 = 0
Find the monic polynomial , in x,  whose roots are 1/r1, 1/r2, 1/r3 and 1/r4 

\(\begin{array}{|rcll|} \hline \mathbf{x^4 - 2x^3 - 5x^2 + 4x - 1} &=& \mathbf{0} \\\\ r^4 - 2r^3 - 5r^2 + 4r - 1 &=& 0 \quad & | \quad : r^4 \\ 1 - 2\dfrac{1}{r} - 5\dfrac{1}{r^2} + 4\dfrac{1}{r^3} - \dfrac{1}{r^4} &=& 0 \quad & | \quad x= \dfrac{1}{r} \\ 1 - 2x - 5x^2 + 4x^3 - x^4 &=& 0 \quad & | \quad *(-1) \\ \mathbf{x^4 -4x^3 + 5x^2+2x-1} &=& \mathbf{0} \\ \hline \end{array}\)

\(\mathbf{x^4 - 2x^3 - 5x^2 + 4x - 1=0} \\ \text{Real solutions}: \\ x=-1.89614823140755 \\ x=3.20188000352314 \\ \text{Complex solutions}: \\ x=0.347134113942206 - 0.210259195466214 i \\ x=0.347134113942206 + 0.210259195466214 i\)

\(\mathbf{x^4 - 4x^3 + 5x^2 + 2x - 1 = 0} \\ \text{Real solutions}: \\ x=-0.527384928792028 \\ x=0.312316513704344 \\ \text{Complex solutions}: \\ x=2.10753420754384 + 1.27653384988106 i \\ x=2.10753420754384 - 1.27653384988106 i\)

someone answered this question on reddit

https://www.reddit.com/r/cheatatmathhomework/comments/f46ajg/pre_calctrig_please_and_thank_you/

Help Please! Find the residue of \(182\cdot 12 - 15\cdot 7 + 3\pmod{14}\).

\(\begin{array}{|rcll|} \hline && \mathbf{182\cdot 12 - 15\cdot 7 + 3\pmod{14}} \\ &=& 2082 \pmod{14} \\ &=& 148*14 + 10 \pmod{14} \\ &=& 0 + 10 \pmod{14} \\ &=& \mathbf{ 10 \pmod{14} } \\ \hline \end{array}\)

Question

\(\begin{array}{|rcll|} \hline \mathbf{f(n)} &=& \mathbf{\log_{10}(2^n)} \\ \hline \\ && \mathbf{\dfrac{f(n)}{\log_{10}(2)}} \\\\ &=& \dfrac{\log_{10}(2^n)}{\log_{10}(2)} \\\\ &=&n \dfrac{\log_{10}(2)}{\log_{10}(2)} \\\\ &=& \mathbf{n} \\ \hline \end{array} \)

The horizontal asymptote of f(g(x)) is y = -5/4.

We can substitute b = 9a to get a^(9a) = (9a)^a.  Then a = 9^(1/3), so m + n = 3 + 9 = 12.

The probability is 5/(3*pi), so a - b = 5 - 3 = 2.

The minimum vaue is 3.

The length of the hallway is 1*90/2.5 = 36 meters.

Let u, v, and w be vectors satisfying \(\mathbf{u}\bullet \mathbf{v} = 3,\ \mathbf{u} \bullet \mathbf{w} = 4,\ \mathbf{v} \bullet \mathbf{w} = 5\).

Then what are \((\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w},\ (\mathbf{w} - \mathbf{u})\bullet \mathbf{v},\ (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u}\) equal to?

\(\begin{array}{|rcll|} \hline \mathbf{u}\bullet \mathbf{v} = \mathbf{v}\bullet \mathbf{u} &=& 3 \\ \mathbf{u} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{u} &=& 4 \\ \mathbf{v} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{v} &=& 5 \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline && (\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w} \\ &=& \mathbf{u} \bullet \mathbf{w}+2\mathbf{v} \bullet \mathbf{w} \\ &=& 4+2\cdot 5 \\ &=&\mathbf{ 14 } \\ \hline \end{array} \begin{array}{|rcll|} \hline && (\mathbf{w} - \mathbf{u})\bullet \mathbf{v} \\ &=& \mathbf{w} \bullet \mathbf{v} - \mathbf{u}\bullet \mathbf{v} \\ &=& 5-3 \\ &=&\mathbf{ 2 } \\ \hline \end{array} \begin{array}{|rcll|} \hline && (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u} \\ &=& 3\mathbf{v}\bullet \mathbf{u} - 2\mathbf{w} \bullet \mathbf{u} \\ &=& 3 \cdot 3 -2\cdot 4 \\ &=& 9-8 \\ &=&\mathbf{ 1 } \\ \hline \end{array} \)

The area of the rhombus is 2*sqrt(3)/2*2^2 = 2*sqrt(3).

Remember that Pi represents the first column, etc. Thus

\(\mathbf{P} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)

Find the distance between the x-intercept and the y-intercept of the graph of the equation \(2x-5y=10\).

\(\begin{array}{|rcll|} \hline 2x-5y &=& 10 \quad & | \quad : 10 \\\\ \dfrac{2x}{10}-\dfrac{5y}{10} &=& 1 \\\\ \dfrac{ x}{5}-\dfrac{ y}{2} &=& 1 \\\\ \mathbf{\dfrac{ x}{5}+\dfrac{ y}{-2}} &=& \mathbf{1} \\ \hline \end{array} \)

x-intercept \(= 5\)

y-intercept \(= -2\)

The distance:

\(\begin{array}{|rcll|} \hline && \sqrt{5^2+(-2)^2} \\ &=& \sqrt{25+4} \\ &=& \mathbf{\sqrt{29}} \\ \hline \end{array}\)

Equation of first line: y = -1/5*x + 7/5:
Equation of second line: y - 3 = -5(x + 8)

y = -5x - 37.

3y = -15x - 111

15x + 3y = -111

(M,N) = (3,-111)

Wait thats incorrect!

Slope = (1 - 5)/(6 -(-3)) = -4/9

point-slope form: y - 5 = -4/9(x + 3)

==> 9y - 45 = -4(x + 3) = -4x - 12

==> 4x + 9y = 33.

A + B + C = 4 + 9 + 33 = 46