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Ierofanga [76]
4 years ago
14

How do you solve -1/3y-6=-11?

Mathematics
2 answers:
Ket [755]4 years ago
5 0

Answer:

y=15

Step-by-step explanation:

-1/3(15)-6=-11

-5-6=-11

lys-0071 [83]4 years ago
4 0

hm

I was wrong look away

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Graph the line that passes through the points (9,-8) and (9,6) and determine the
avanturin [10]

Answer:

<em>x = 9</em>

Step-by-step explanation:

The standard form of equation of a line is expressed as y = mx+c

m is the slope

c is the y intercept

Note that the line does not have a slope and the x intercept of the line is a point where the line cuts the x axis. Hence the equation of the line will be expressed as x = c where c is the x intercept;

<em>From the graph, the x intercept =9. Hence the equation of the given line is x = 9</em>

7 0
3 years ago
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}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\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}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

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

\bf sin(\theta)=\cfrac{-1}{6}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{\sqrt{35}}{6}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{-1}{\sqrt{35}}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{\sqrt{35}}{1}&#10;\\\\\\&#10;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}&#10;\\\\\\&#10;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
4 years ago
On a piece of paper, use a protractor to construct a triangle with angle measures of 40 degrees and 60 degrees.
Andru [333]

Answer:

80

Step-by-step explanation

Well every triangle has to equal 180 degrees.

So you would add the two degrees you already have, which on this one you would get 100.

Then, you would subtract that number from 180 to get 80.


Hope this helps you!!

8 0
3 years ago
Read 2 more answers
Help please don’t answer if you don’t know also I’ll give brainliest
egoroff_w [7]

Answer:

10x=4

x=4/10

x=2/5

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Answers? Or a hint please I’m begging
olga2289 [7]
Hmm I would say x=q
Hope this helps u <3
6 0
3 years ago
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