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Nataly_w [17]
1 year ago
6

Divide. Assume that all variables in the denominator are nonzero.

Mathematics
2 answers:
Genrish500 [490]1 year ago
6 0

Answer:

4a^{5}

Step-by-step explanation:

\frac{64}{16} = 4

When the bases are the same and you are dividing you subtract the exponents.  It is easier to see why when you write the problem out in expanded form.

\frac{aaaaaabbbbbbb}{abbbbbbb}  If you cross out all of the a's on top with the a's on the bottom, you are left with only 1 a (6-1 = 5)  If you do not see an exponent, it is understood to be 1.  All of the b's on the top and bottom cross out. (7-7 = 0)  Anything to the zero power = 1.

yarga [219]1 year ago
4 0

Answer:

4a^{5}

Step-by-step explanation:

The easiest way to start this solving is simplifying the numbers:

start by dividing 64 by 16. This gets you 4.

Then simplify the variables. Let's start with <em>a</em>:

to do this, we will start by moving all the <em>a'</em>s to the same line so we can multiply them

multiplying like variables means that you are actually just adding their exponents

when moving a variable up or down on the fraction, the exponent will flip signs (positive to negative, or negative to positive)

in this case, we have a^{6} in the numerator and a^{1}, so it would be easiest to move   a^{1} above so that it the equation will become \frac{4a^{6}a^{-1}b^{7}   }{b^{7} }

now we can subtract 1 from 6 and get  \frac{4a^{5}b^{7}   }{b^{7} }

Now let's work on simplifying <em>b:</em>

both the numerator and denominator have b^{7} in it

so if we move the <em>b</em> denominator up, the numerator ends up as     {4a^{5}b^{7}b^{-7} }

which makes both <em>b'</em>s cancel out, making your final answer 4a^{5}

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2a. 14,9; 2b. 15,4; 2c. 30,8; 3. 32

For 2a., you have to set it up like this: csc 48° = 20⁄x [OR sin 48° = ˣ⁄20]. Then you would have to isolate the variable by getting rid of the denominator. The cosecant function has an extra step because you will get xcsc 48° = 20. As stated, isolate the variable; this time, divide by csc 48°. This is what 20 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2b., you have to set it up like this: sec 39° = ˣ⁄12 [OR cos 39° = 12⁄x. Then you would have to isolate the variable by getting rid of the denominator. The cosine function has an extra step because you will get xcos 39° = 12. As stated, isolate the variable; this time, divide by cos 39°. This is what 12 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2c., you have to set it up like this: cot 64° = 15⁄x [OR tan 64° = ˣ⁄15]. Then you would have to isolate the variable by getting rid of the denominator. The cotangent function has an extra step because you will get xcot 64° = 15. As stated, isolate the variable; this time, divide by cot 64°. This is what 15 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

Now, for 3., it is unique, but similar concept. In this exercise, we are solving for an angle measure, so we have to use inverse trigonometric ratios. So, we set it up like this: cot⁻¹ 1⅗ = m<x [OR tan⁻¹ ⅝ = m<x]. We simply input this into our calculator and we get 32,00538321°. When rounded to the nearest degree, we get 32°.

WARNING: If you use a graphing calculator, you have to input it uniquely because most graphing calculators do not have the inverse trigonometric ratios programmed in their systems. This is how you would write this: tan⁻¹ 1⅗⁻¹. You set 1⅗ to the negative first power, ALONG WITH the inverse tangent function, because without it, your answer will be thrown off. Since Cotangent and Tangent are multiplicative inverses of each other, that is the reason why the negative first power is applied ALONG WITH the inverse tangent function.

**NOTE: 1⅗ = 8⁄5

Take into consideration:

sin <em>θ</em> = O\H

cos <em>θ</em><em> </em>=<em> </em>A\H

tan <em>θ</em><em> </em>= O\A

sec <em>θ</em><em> </em>= H\A

csc <em>θ</em><em> </em>= H\O

cot <em>θ</em><em> </em>= A\O

I hope this helps you out alot, but if you are still in need of assistance, do not hesitate to let me know and subscribe to my channel [username: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

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