Mathematics, 03.05.2021 07:30 chivitogomez2400

Trigonometry sine question: Based on this diagram and the right triangle definition of the sine ratio, find an expression for sin q in terms of the variables x, y and r. sin(Ф) = ?

Trigonometry sine question: Based on this diagram and the right triangle definition of the sine rat

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Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

Answer from: Quest

25% is the answer

Answer from: Quest

so, converting firstly the mixed fractions to improper fractions and then "adding".

$\bf \stackrel{mixed}{2\frac{9}{10}}\implies \cfrac{2\cdot 10+9}{10}\implies \stackrel{improper}{\cfrac{29}{10}}~\hfill \stackrel{mixed}{5\frac{4}{15}}\implies \cfrac{5\cdot 15+4}{15}\implies \stackrel{improper}{\cfrac{79}{15}} [-0.35em] ~\dotfill$

$\bf -\cfrac{29}{10}+\left( -\cfrac{79}{15} \right)\implies -\cfrac{29}{10}-\cfrac{79}{15}\implies \stackrel{\textit{using the lcd of 30}}{\cfrac{-(3)29-(2)79}{30}}\implies \cfrac{-87-158}{30} \cfrac{-245}{30}\implies -\cfrac{49}{6}\implies -8\frac{1}{6}$

Answer from: Quest

a. we have two

lines: y = 4-x and

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b. 4-x = 8-x^-1

x 4-x 8-x^-1

-3 7

8.33

-2 6 8.5

-1 5 9

0 4

1 3

2 2

7.5

3 1 7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:

x=-4.236

y = 4-x = 4 – (-4.236)

= 8.236

y = 8-x^-1 = .236)^-1

= 8.236

x=0.236

y = 4-x = 4 – (0.236) = 3.764

y = 8-x^-1 = 8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c. to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x and y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines. the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

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