Can someone please help me

Mathematics

1 answer:

Maurinko [17]4 years ago

6 0

This isn't an answer, rather a suggestion.

I would use the website desmos to help you out. But it won't help you forever like when you do quizzes or tests.

Directions:

Make the variable "y" all by itself. It need to be solo. For example for problem 34, negate 3 on each side and you will have "y=1/2(x+2)+3".

From there, plug in numbers into the x variable. (I suggest using common factors of "2" since the number your multiplying is 1/2.

Again using 34,

y=1/2(x+2)+3

y=1/2(2+2)+3

y=1/2(4)+3

y=2+3

y=5

The coordinates are (2,5). Plot the coordinate.

Repeat the process.

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Digiron [165]

Given:

<span>The average number of daylight hours each month in Springfield in 2015:

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5 0

3 years ago

The angle θ lies in Quadrant II .

Andreyy89

let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.

cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}$