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2 years ago

9

The state of Oregan has a population of 20 people per square mile. A county in oregan measures 25 miles by 30 miles. What is the

expected population of the county

1 answer:

Juli2301 [7.4K]2 years ago

8 0

The answer is 15.000 I hope this helps your question;)

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Write an absolute value function that is reflected over the x axis, moved left 2 units and up 5 units

MrMuchimi

Answer:

I really don't know i don't care

Step-by-step explanation:

plz plzplzplzp you can u just give 500fb a 75hz to get amazon for free and you are you for the only reason 5-pack 3fE6bFt 5-pack 3d is not an average annual 75hz 8

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2 years ago

(a) An angle measures 43 . What is the measure of its complement? (b) An angle measures 81 . What is the measure of its suppleme

vampirchik [111]

To solve the problem shown above, you must follow the proccedure shown below:

1. By definition, Completary angles are those angles whose sum is 90 degrees and Suplementary angles are those angles whose sum is 180 degrees.

2. Keeping the information above on mind, you have:
<span>
(a) An angle measures 43 . What is the measure of its complement?

=90°-43°
=47°

(b) An angle measures 81 . What is the measure of its supplement?
</span>
=180°-81°
=99°

The answers are:

a) 47°
b) 99°

7 0

2 years ago

{(-4,8),(-2,4),(0,1),(2,4),(4,8)}

astraxan [27]

Answer:

Since there is one value of y for every value of x in (−4,8),(−2,4),(0,1),(2,4), ( 4,8), this relation is a function. The relation is a function.

5 0

2 years ago

Esther starts at one corner of a square field and walks 90 feet along one side to another corner of the field. She turns 90° and

erma4kov [3.2K]

The three sides make a right triangle and you know the base (90) and the height (70)

A = 1/2 bh or 1/2 x (70)x (90)= 3150 ft²

3 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

——————————

Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago