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I am Lyosha [343]
3 years ago
9

.

Mathematics
1 answer:
Sholpan [36]3 years ago
7 0

Answer:

(81pi)/80

Step-by-step explanation:

i just got this wrong

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Step-by-step explanation:

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Divide and simplify. Give your answer using positive exponent. Leave your answer in exponential notation.
yaroslaw [1]

Answer:

14⁻⁵

Step-by-step explanation:

Dividing two exponents with the same base has the effect of <em>subtracting</em> the bottom exponent from the top one. Following that rule, we can simplify our fraction like this:

\dfrac{14^{-3}}{14^2}=14^{-3-2}=14^{-5}

3 0
3 years ago
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What is the length of line segment RS? Use the law of sines to find the answer. Round to the nearest tenth. Law of sines:
Leno4ka [110]

Answer:

Correct Answer is B. 2.4 Units

Step-by-step explanation:

The missing picture in question is attached.

The Law of Sines is given as:

\frac{Sin(A)}{a} = \frac{Sin (B)}{b} = \frac{Sin (C)}{c}

Where,

a,b,c are the length of sides of triangle

A,B,C are the angles between the two sides of triangle

According to the picture attached, we have ΔRQS,

Let,

r = 3.1 units

R = 80°

s = 2.4 units

S = S

q = RS

Q = Q

Using law of Sines:

\frac{Sin (R)}{r} = \frac{Sin (S)}{s}\\ \frac{Sin (80)}{3.1} = \frac{Sin (S)}{2.4} \\Sin (S) = \frac{Sin (80)}{3.1} * 2.4\\Sin (S) = 0.762\\S = Sin^{-1} (0.762) \\S = 49.68°

Since, triangle is constitute of total 180°, hence,

∠Q + ∠R + ∠S = 180°

∠Q + 80 + 49.68 = 180

∠Q = 180 - 80 - 49.68

∠Q = 50.32°

To find line segment RS, again use law of sines:\frac{Sin (R)}{r} = \frac{Sin (Q)}{q}\\\frac{Sin (80)}{3.1} = \frac{Sin (50.32)}{RS}\\0.3177 = \frac{0.77}{RS}\\RS = 2.42 Units

7 0
3 years ago
Read 2 more answers
If cos Θ = negative 4 over 7, what are the values of sin Θ and tan Θ? (2 points)
seropon [69]
I accept that θ in second quadrant, therefore sinθ > 0, cosθ < 0 and tanθ < 0.

\cos\theta=-\dfrac{4}{7}\\\\We\ know:\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\to\sin x=\sqrt{1-\cos^2x}\\\\subtitute:\\\\\sin x=\sqrt{1-\left(-\dfrac{4}{7}\right)^2}=\sqrt{1-\dfrac{16}{49}}=\sqrt{\dfrac{33}{49}}=\dfrac{\sqrt{33}}{7}

We\ know:\tan x=\dfrac{\sin x}{\cos x}\\\\subtitute:\\\\\tan x=\dfrac{\sqrt{33}}{7}:\left(-\dfrac{4}{7}\right)=-\dfrac{\sqrt{33}}{7}\cdot\dfrac{7}{4}=-\dfrac{\sqrt{33}}{4}
5 0
3 years ago
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