All categories

2 years ago

Refer to the article "Fire Up the Sun."

1 answer:

8 0

Based on the information given, it should be noted the central idea is C. Making and using a solar oven is a learning experience.

<h3>What is a central idea?</h3>

A central idea simply means the main idea that's in a literary work. It's the unifying element that's in a story.

The central idea is that making and using a solar oven is a learning experience and the detail that supports the central idea is that "by completing this project, you will make an inexpensive, ecologically friendly piece of equipment and learn about the science behind what you’re doing".

Learn more about central ideas on:

brainly.com/question/2684713

You might be interested in

Solve the following triangle

MArishka [77]

Answer:

Step-by-step explanation:

as , 77+x = 146(external angle) , so x = 146 - 77

so x = 69

4 0

3 years ago

Read 2 more answers

I need help finding Sin B, Cos B, and Tan A for a geometry problem. please help meee

Scilla [17]

Answer:

Sin B = 5/13

Cos B = 12/13

tan A = 12/5

Step-by-step explanation:

Sin B = opposite side/ hypotenuse

Sin B = 5/13

Cos B = adjacent side / hypotenuse

= 12/13

tan A = opposite side /adjacent side

= 12/5

6 0

3 years ago

PLEASE HELP. zoom in to remove blur

AleksAgata [21]

Answer:

It's A

Step-by-step explanation:

It's just moving the equation to the right 4 units. There's no stretch, compression, or reflection.

5 0

2 years ago

Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0

3 years ago

What is the length of BC , rounded to the nearest tenth?

Arte-miy333 [17]

Step $1$