Write the equation of the line that passes through the points (-9, -7) and (-4, 4).

Mathematics

1 answer:

Amanda [17]3 years ago

6 0

Answer:

$y-4 = \frac{11}{5} (x+4)$

Step-by-step explanation:

1) First, find the slope of the line. Use the slope formula $m = \frac{y_2-y_1}{x_2-x_1}$ to find the slope. Substitute the x and y values of the given points into the formula and solve:

$m = \frac{(4)-(-7)}{(-4)-(-9)} \\m = \frac{4+7}{-4+9} \\m = \frac{11}{5} \\$

Thus, the slope is $\frac{11}{5}$ .

2) Now, use the point-slope formula, or $y-y_1 = m (x-x_1)$ to write the equation of the line in point-slope form. Substitute values for $m$ , $x_1$ , and $y_1$ .

Since $m$ represents the slope, substitute $\frac{11}{5}$ for it. Since $x_1$ and $y_1$ represent the x and y values of one point on the line, choose any one of the given points (either one is fine, the equation will represent the same line) and substitute its x and y values into the formula as well. (I chose (-4, 4) as seen below.) Make sure to simplify the double negatives to positives. This gives the following answer in point-slope form:

$y-(4) = \frac{11}{5} (x-(-4))\\y-4 = \frac{11}{5} (x+4)$

You might be interested in

Plz help I will make you brainliest!Thanks

STatiana [176]

Use the formula V=AbH and 48=16h will be the second step and then divide by 16 and then you get the answer

7 0

3 years ago

Read 2 more answers

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,

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

3 0

3 years ago

I NEED HELP ON THIS QUESTION ASAP

Fittoniya [83]

Answer:

(2.5, -.25)

Step-by-step explanation:

I'm not positive that I know exactly what they want on this one....but since you specifically asked me to look at your other question, I tried.

See pic.