Okay first you have to divide
2 goes into 3
3/2
2 fits into 3 once
1 and the left over is 1 so
1 1/2
Answer:
10cos(5x)sin(10x) = 5[sin (15x) + sin (5x)]
Step-by-step explanation:
In this question, we are tasked with writing the product as a sum.
To do this, we shall be using the sum to product formula below;
cosαsinβ = 1/2[ sin(α + β) - sin(α - β)]
From the question, we can say α= 5x and β= 10x
Plugging these values into the equation, we have
10cos(5x)sin(10x) = (10) × 1/2[sin (5x + 10x) - sin(5x - 10x)]
= 5[sin (15x) - sin (-5x)]
We apply odd identity i.e sin(-x) = -sinx
Thus applying same to sin(-5x)
sin(-5x) = -sin(5x)
Thus;
5[sin (15x) - sin (-5x)] = 5[sin (15x) -(-sin(5x))]
= 5[sin (15x) + sin (5x)]
Hence, 10cos(5x)sin(10x) = 5[sin (15x) + sin (5x)]
Answer:
r > 17
Step-by-step explanation:
r -7> 10
Add 7 to each side
r-7+7 > 10+7
r > 17
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Answer:</u></em></h2><h2><em><u>
Sorry i didn't understand the question but thank you for the points </u></em></h2><h2><em><u>
Step-by-step explanation:</u></em></h2>
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The summation of negative 5 n minus 1, from n equals 3 to 12 can be
expressed as
n = 3/12
substitute the value of n to the equatio
-5n – 1
-5( 3/12 ) -1
-9/4
