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Dmitry_Shevchenko [17]
3 years ago
6

Solve for w and simplify this fraction 3/2w=-20​

Mathematics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

If I'm correct, the answer should be  negative 40/3.

Step-by-step explanation:

Because you first combine multiplied terms into a single fraction, then you multiply all terms by the same value to eliminate fraction denominators. Now you simplify it, by canceling multiplied terms that are in the denominator and then you multiply the numbers. Now you will divide both sides of the equation by the same term. Finally, you will simplify the equation.

(Hope this helps) I had to get a little help myself on this problem)

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Which set of ordered pairs represents a function?
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Answer:

the second one

Step-by-step explanation:

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Fofino [41]

Answer:  see proof below

<u>Step-by-step explanation:</u>

\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}

\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}

\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}

\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}

\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}

\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark

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