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3 years ago

Find x when y=5 if y varies inversely as x and x=6 when y =-18

1 answer:

7 0

It has been given that y varies inversely with x, therefore the equation for this relationship is as follows;
y = $\frac{k}{x}$ \
when y = -18 and x = 6
-18 = k/ 6 ---1)
We need to find x when y = 5
5 = k/x
when we divide 1)/2), k gets cancelled

$\frac{-18}{5}= \frac{x}{6}$
x = -18*6/5
x = -21.6
when y = 5, then x = -21.6

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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,

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

3 0

3 years ago

Find the slope (-2,-4) (2,4)

Alex

Answer:

8 over 4

Step-by-step explanation:

5 0

3 years ago

What is W in this equation -w–11≤w–5

zhannawk [14.2K]

Step-by-step explanation:

-w - 11 < or = w - 5

-2w - 11 < or = -5

-2w < or = -16

w > or = -8

5 0

3 years ago

Which of the following pairs of functions are inverses of each other?

Serggg [28]

Answer:

The answer is (B).

Step-by-step explanation:

A). y = f(x) = 6(x - 2) + 3 ⇔ y = 6x - 12 + 3 ⇔ 6x = y + 9 ⇔ x = $\frac{y+9}{6}$

g(x) = $\frac{x+9}{6}$ is inverse function of f(x) and ≠ $\frac{x+2}{6}$ - 3

B). y = f(x) = $\frac{5x}{4}$ - 3 ⇔ 4y = 5x - 12 ⇔ x = $\frac{4y+12}{5}$

g(x) = $\frac{4(x + 3)}{5}$ is inverse of f(x)

6 0

3 years ago

Identify the type of conic section that is represented by the equation y^2- 1 = 4(x-7). DUE IN 2 HOURS, will give BRAINLIEST! A.

Elenna [48]

Answer:

Solution : Parabola

Step-by-step explanation:

As you can see only one variable is square in this situation, so it can only be a parabola. We can prove that it is a parabola however by converting it into standard form (x - h)^2 + (y - k)^2.

$y^2-1=4\left(x-7\right) = y^2-1=4\left(x-7\right)$

Respectively it's properties would be as follows,

$\left(h,\:k\right)=\left(\frac{27}{4},\:0\right),\:p=1$

8 0

3 years ago