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Mazyrski [523]
2 years ago
10

Sin 2A + sin A/1+cos A + cos 2A=tan A

Mathematics
1 answer:
Nezavi [6.7K]2 years ago
7 0

Answer:

= 2sinA.cosA+ sinA / 1+cos2A+ cosA

= sinA(2cosA + 1) / 2cos^2A + cosA

= sinA(2cosA + 1) / cosA( 2cosA + 1)

= sinA / cosA

= tan A

= R.H.S

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Answer:

You would want to do the square root of 361 which is 19

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5 0
2 years ago
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Please solve for M<PRQ​
MakcuM [25]

Answer:

m\angle PRQ =49\degree

Step-by-step explanation:

m\angle PSQ = 360\degree - (128\degree + 134\degree )

m\angle PSQ = 360\degree - 262\degree

m\angle PSQ = 98\degree

Now, by inscribed angle theorem:

m\angle PRQ = \frac{1}{2} \times m\angle PSQ

m\angle PRQ = \frac{1}{2} \times 98\degree

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5 0
2 years ago
Evaluate the expression when k=-4<br> |k-2 • |3k|,- 1<br> a. 72<br> b. 71<br> c. 66<br> d. -73
Feliz [49]

Answer:

-6×12×-1=72

Step-by-step explanation:

just note that abs(k)

when k<0

is -k

6 0
3 years ago
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LOTS OF POINTS!! A cyclist rides a 25 mile stretch of road at a constant speed. It takes her 90 minutes to ride the 25 miles . A
sattari [20]
25m/90minutes = .278 mile/minute

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3 years ago
If a/b &lt; c/d with b &gt; 0, d &gt; 0, prove that a+c / b+d lies between the two fractions a/b and c/d .​
Tems11 [23]

Divide through everything by <em>b</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}

Since <em>a/b</em> < <em>c/d</em>, it follows that

\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}

Multiply through everything on the right side by <em>b/d</em> to get

\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}

and <em>a/b</em> < <em>c/d</em> tells us that

\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}

Then

\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

5 0
3 years ago
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