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ZanzabumX [31]
3 years ago
12

2/5x = -3/15 + 1/x x=0 x=15 x=5 x=3

Mathematics
1 answer:
Taya2010 [7]3 years ago
7 0

Answer:

x = 3

Step-by-step explanation:

\dfrac{2}{5x}=-\dfrac{3}{15}+\dfrac{1}{x}\\\\\dfrac{1}{5}=\dfrac{5-2}{5x}\qquad\text{add $\frac{3}{15}-\frac{2}{5x}$}\\\\x=3\qquad\text{multiply by 5x}

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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
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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{}

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3 years ago
Will mark brainliest :)
fiasKO [112]

Answer:

B im pretty sure

Step-by-step explanation:

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Which number(s) below belong to the solution set of the inequality? Check all that apply.x + 16 < 51.
natta225 [31]
If you subtract 16 on both sides... x will be alone. Therefore 51-16=35
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Solve for 2x+3y+6z when x=4, y=3, and z=6.
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Answer:

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Step-by-step explanation:

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