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2 years ago

Write the ratios for sin x, cos x and tan x

Mathematics

1 answer:

yuradex [85]2 years ago

4 0

Answer:

sin(x)= $\sqrt{119\\}$ / 12, cos(x)=5/12, tan= $\sqrt{119\\}$ / 5

Step-by-step explanation:

segment YZ is length a, we need to find a

segment XZ is length 5

segment XY is length 12

XYZ is a right triangle, so a^2 + b^2 = z^2, so

a^2 + 5^2 = 12^2, simplify

a^2 + 25 = 144, subtract 25

a^2 = 119

a = $\sqrt{119\\}$

then we can determine all the trig ratios

soh - cah - toa

sin = opp/hyp, cos=adj/hyp, tan=opp/adj

sin(x)= $\sqrt{119\\}$ / 12, cos(x)=5/12, tan= $\sqrt{119\\}$ / 5

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Write the equation of the line in fully simplified slope-intercept form.

Lina20 [59]

Answer:

$y=-6x+2$

Step-by-step explanation:

Hi there!

Slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when x=0)

1) Determine the slope (m)

$m=\displaystyle \frac{y_2-y_1}{x_2-x_1}$ where two points that fall on the line are $(x_1,y_1)$ and $(x_2,y_2)$

Given the graph, we determine which points we could use. For example, we could use the two points $(0,2)$ and $(1,-4)$ :

$m=\displaystyle \frac{-4-2}{1-0}\\\\m=\displaystyle \frac{-6}{1}\\\\m=-6$

Therefore, the slope of the line is -6. Plug this into $y=mx+b$ :

$y=-6x+b$

2) Determine the y-intercept (b)

Recall that the y-intercept occurs when x=0. Given the point (0,2), we know that the y-intercept is 2. Plug this into $y=-6x+b$ :

$y=-6x+2$

I hope this helps!

5 0

3 years ago

On friday margaret biked 1 5/8 times the lengtof the bike path how many milea did she ride?

zaharov [31]

Multiple 15 times 8 and that is your answer

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3 years ago

It is estimated that there are approximately 1 × 1014 cells in the human body. Which of the following represents this amount in

serg [7]

1014 hope thts the awnser it seemed to obvious

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3 years ago

Which of the following is an extraneous solution of (45-3x)^1/2=x-9?

Iteru [2.4K]

Answer:

C. x =3

Step-by-step explanation:

Extraneous solution is that root of a transformed equation that doesn't satisfy the equation in it's original form because it was excluded from the domain of the original equation.

Let's solve the equation first

$\sqrt{45-3x} = x-9\\Taking\ square\ on\ both\ sides\\{(\sqrt{45-3x})}^2 = {(x-9)}^2\\45-3x = x^2-18x+81\\0 = x^2-18x+81-45+3x\\x^2-15x+36 = 0\\x^2-12x-3x+36 = 0\\x(x-12)-3(x-12) = 0\\(x-3)(x-12)\\x-3 = 0\\=> x =3\\x-12 = 0\\x = 12\\We\ will\ check\ the\ solutions\ one\ by\ one\\So,\\for\ x=3\\\sqrt{45-3(3)} = 3-9\\\sqrt{45-9} = -6\\\sqrt{36}= -6\\6\neq -6\\For x=12\\\sqrt{45-3(12)} = 12-9\\\sqrt{45-36} = 6\\\sqrt{36}= 6\\6=6$