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2 years ago

Find the slope of the line that passes through the points (1, 7) and (3, 3).

Mathematics

2 answers:

Alenkasestr [34]2 years ago

7 0

Answer:

Slope of the line that passes through the points (1, 7) and (3, 3) = -2

Step-by-step explanation:

Given in the question two co-ordinates,

(1, 7)

(3, 3)

here x1 = 1

x2 = 3

y1 = 7

y2 = 3

Formula to use

m = $\frac{y2 - y1}{x2 - x1}$

= $\frac{3-7}{3-1}$

= -4/2

= -2

Eddi Din [679]2 years ago

6 0

Answer:

The slope of given line is -2

Step-by-step explanation:

<u>Points to remember</u>

Let (x₁, y₁) and (x₂, y₂) be the two points in a line, then the slope of the line is given by,

slope, m = (y₂ - y₁)/(x₂ - x₁)

<u>To find the slope of given line</u>

Here (x₁, y₁) = (1, 7) and (x₂, y₂) = (3,3)

Slope m = (y₂ - y₁)/(x₂ - x₁) = (3 - 7)/(3 - 1)

= -4/2 = -2

Therefore the slope of given line is -2

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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

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Ksenya-84 [330]

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Step-by-step explanation:

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natali 33 [55]

Answer:

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Step-by-step explanation:

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