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Bogdan [553]
3 years ago
9

Find the missing lengths of the sides.

Mathematics
1 answer:
ra1l [238]3 years ago
4 0
The answer is C, you can solve this by sin(30)= 8/c since sine is opposite over hypotenuse and since c equals 16 there is only one possible solution 
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Juan has $10. He recently got a job and now makes $9.15 an hour. He now has $760.30. How many hours has he worked?
marissa [1.9K]

Im not sure if im correct but I think the answer is 82 hours.

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Help me plz I need help
charle [14.2K]

Answer:

-0.48

Step-by-step explanation:

(-1.2)(0.4)

= - 0.48

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Twice a year, for all four years of college, you used your credit card to pay for books and supplies, which amounts to $516 a se
mixer [17]

Answer:

$11,728

Step-by-step explanation:

Twice a year, for 4 years is 8 times

516×8 = 4128

Yearly for 4 years is 4 times

700×4 = 2800

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3 years ago
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
Please help serious answers only
Dmitry_Shevchenko [17]
It’s 28, 16 add 4 is 20 add 8 is 28
7 0
3 years ago
Read 2 more answers
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