Could someone show me the steps to this problem

Mathematics

1 answer:

Lyrx [107]3 years ago

5 0

You usually just divide, so the answer is 4.07692308

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The distance between two towns is 21.673 km. Round of the distance to nearest 0.01 km.

Volgvan

Answer:

21.67

Step-by-step explanation:

look at the number in the 0.01 value then the number next to it

if it's five or above then the 7 goes up if it's 4 and below then the number stays the same. in this case the number will stay the same

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3 years ago

Find the degree of the term. -9x4

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Degree of the Term is the sum of the exponents of the variables. 2x 4y 3 4 + 3 = 7 7 is the degree of the term. 5x-2y 5 NOT A TERM because it has a negative exponent. 8 If a term consists only of a non-zero number (known as a constant term) its degree is 0. Your welcome!

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The number 78 is divisible by 2 and 3. True False

vladimir1956 [14]

True, the number 78 is divisible by both 2 and 3.

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3 years ago

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Find the exact value of tan(165°) using a difference of two angles

Katyanochek1 [597]

Answer: $-2+\sqrt{3}$

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Work Shown:

Apply the following trig identity

$\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\$

Now let's rationalize the denominator

$\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\$

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As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

$\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949$