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3 years ago

If r=10 and s=31 find R. Round to the nearest tenth

Mathematics

2 answers:

Natasha2012 [34]3 years ago

6 0

Answer:

The correct answer option is C. 17.9°.

Step-by-step explanation:

We are given a right angled triangle, SRT, with two known sides, r and s.

We are to find the measure of the angle R.

For that, we will use tan.

$tan R = \frac { r } { s }$

$tan R = \frac { 1 0 } { 3 1 }$

$R = tan' 0.322$

R = 17.9°

Vsevolod [243]3 years ago

3 0

Answer: option c.

Step-by-step explanation:

You need to remember the identity:

$tan\alpha=\frac{opposite}{adjacent}$

The inverse of the tangent function is arctangent. You need to use this to calculate the angle "R":

$\alpha =arctan(\frac{opposite}{adjacent})$

You know that you need to find the measure of "R" and $r=10$ (which is the opposite side) and $s=31$ (which is the adjacent side), you can sustitute values into $\alpha =arctan(\frac{opposite}{adjacent})$

Then, you get:

$R=arctan(\frac{10}{31})\\\\R=17.9\°$

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Put in order from least to greatest

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Answer

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3 years ago

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3 years ago

Lim x-1 x2 - 1/ sin(x-2)

balu736 [363]

Answer:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0$

Explanation:

Assuming the correct expression is to find the following limit:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}$

Evaluate the numerator:

$\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}$

Evaluate the denominator:

Since $\lim_{x \to1}sin(x-2)\neq 0$

$\frac{0}{\lim_{x \to1}sin(x-2)}=0$

4 0

3 years ago

Simplify 5^2 • 5^4 O A. 5^8 O B. 5^6 O C. 5•8 O D. 5^2

amid [387]

Answer:

Option B

Step-by-step explanation:

Given the following question:

$5^2\times5^4$

In order to find the answer, we simply apply the exponent rule and add the exponents.

$5^2\times5^4$
$2+4=6$
$=5^6$

The answer is option B or "5^6."

Hope this helps.