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IrinaVladis [17]
3 years ago
11

HELP WILL MARK YOU BRAINLIEST!!!

Mathematics
1 answer:
Vlad [161]3 years ago
3 0

Answer:

<PQR=132°

plz mark me brainliest

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5/6k=-20 what is the correct answer ??
damaskus [11]
<span>5/6k=-20

k=-24

Hope this helps
</span>
4 0
3 years ago
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What is the simplified form of 2(3x+5)-30+7x_4
Kipish [7]

Answer:

The answer is 6x+7x_4-20

Step-by-step explanation:

6 0
3 years ago
Angles 1 and 2 are supplementary.
Phantasy [73]

Answer:

m∡1 + m∡2 = 180

Step-by-step explanation:

m∡1 + m∡2 = 180

This is because supplementary means they add up to 180°.

3 0
3 years ago
What is the value of x in the equation 8 + x = 3?<br><br> A. −5<br> B. 5<br> C. 11<br> D. 24
zlopas [31]

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I think it's A

I hope this helps

3 0
3 years ago
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A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form
Over [174]

Answer:

\vec{v}= \text{ or } \approx

Step-by-step explanation:

Component form of a vector is given by \vec{v}=, where i represents change in x-value and j represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector \vec{v}=, the magnitude is ||v||=\sqrt{i^2+j^2.

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being i, one leg being j, and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}

To find the other leg, i, we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}

Verify that (9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark

Therefore, the component form of this vector is \vec{v}=\boxed{}\approx \boxed{}

6 0
3 years ago
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