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maxonik [38]
3 years ago
8

How many points do you need in order to name a ray?

Mathematics
2 answers:
Kamila [148]3 years ago
7 0
You only need one point to make a ray.
patriot [66]3 years ago
7 0

Answer:

How many points do you need in order to name a ray?


<h2>Your Answer Will Be Two . </h2>

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sukhopar [10]

Answer:

Choose the first one and second and the last one.

Step-by-step explanation:

This is an simple

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3a+6b-a-9b please help!
faust18 [17]
Answer: 2a - 3b
You can try rearranging it , it’ll be easier I think :)
3a + 6b - a - 9b
= 3a - a + 6b - 9b
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10/12 (ten twelves) is an equivalent fraction to 5/7 (five seventh)? True or False
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That is False. Please brainiest me!
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PLS HELP!!! YOU'LL BE A GRADE SAVER. I'LL IVE BRAINLIEST IF CORRECT. Triangle XYZ was dilated by a scale factor of 2 to create t
tankabanditka [31]

Answer:

Step-by-step explanation:

As for the angles of both triangles; they’re the same. The sides are 1:2.

I’m giving you formulas that are labeled: side a shortest

” b mid length

” c hypotenuse

angle α(alpha) opposite side a

” β(beta) ” ” b

” γ(gamma) ” ” c

A major formula for rt triangles is: a^2+b^2=c^2.

*Another is: a/sinα=b/sinβ=c/sinγ.

Remember α+β+γ=180°.

As for sides a&b use the above formula.

As for <ACB; the angle is γ which is a rt <.

Given: tan<x=5/2+1/2=6/2=3atan=71.565……….°=β. So α=18.44…….°. γ= rt angle.

To get the sides use the formulas at *.

6 0
3 years ago
Find the exact value of cos(theta) for an angle (theta) with tan (theta)= -2/3 and with its terminal side in Quadrant II.
Naya [18.7K]

Answer:

-\frac{3\sqrt{13}}{13}.

Step-by-step explanation:

Since we are in quadrant two, cosine value is negative while sine value is positive.

We are going to use the Pythagorean Identity: 1+\tan^2(\theta)=\sec^2(\theta).

1+(\frac{-2}{3})^2=\sec^2(\theta)

1+\frac{4}{9}=\sec^2(\theta)

\frac{9+4}{9}=\sec^2(\theta)

\frac{13}{9}=\sec^2(\theta)

\pm \sqrt{\frac{13}{9}}=\sec(\theta)

\pm \frac{\sqrt{13}}{3}=\sec(\theta)

Since cosine and secant are reciprocals then they will have the same sign as along as they both exist.

\sec(\theta)=-\frac{\sqrt{13}}{3}

\cos(\theta)=-\frac{3}{\sqrt{13}}.

I don't see this answer as I'm going to rationalize the denominator.

\cos(\theta)=-\frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}.

\cos(\theta)=-\frac{3\sqrt{13}}{13}.

8 0
4 years ago
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