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3 years ago

Which relationship describes angles 1 and 2?

Mathematics

2 answers:

mel-nik [20]3 years ago

8 0

<span>adjacent angles
</span><span>complementary angles
</span>this are the answers

katovenus [111]3 years ago

4 0

Answer: The answer is (A) complementary angles.

Step-by-step explanation: We are given a figure involving angles 1 and 2 and we are to select the correct relation between these two angles.

We have

$\angle 1+\angle 2+90^\circ=180^\circ,~\textup{measure of a straight angle is 180 degrees}\\\\\Rightarrow \angle 1+\angle 2=180^\circ-90^\circ\\\\\Rightarrow \angle 1+\angle 2=90^\circ.$

We know that, if the sum of two angles is 90°, then they are said to be complementary angles.

Thus, ∠1 and ∠2 are complementary angles.

Thus., (A) is the correct option.

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J u l i a n has partially filled the p r i s m below with 16 unit cubes. How many more unit cubes will he need to completely fil

Wittaler [7]

add it

Step-by-step explanation:

11+20+25+36

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2 years ago

At his birthday party Joey got $100 from each of his 2 grandmothers $40 from his dad and $5 from sister, how much did he get? Dr

horsena [70]

He got $245 for his birthday

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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

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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

For the partially complete factorization, find the other binomial which will complete the factorization n^2-8n-20=(n+2)(_____)

Otrada [13]

Answer:

(n-10)

Step-by-step explanation:

Jdjdjdhbejdudjkdff

8 0

2 years ago

Please help. !!!!!!!!

Kryger [21]

Basically you find the data points and fill in the values. For Point J it is (2,4). They want you to do this: (2-4,4-6)...it then becomes (-2,-2).
So
J: (-2,-2)
K: (-1, -4)
I: (-2, -5)
Those are your new points. I hope this helps love! :)

8 0

3 years ago