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Kruka [31]
3 years ago
8

I NEED HELP ASAP PLZZZZ HURRRY

Mathematics
1 answer:
AleksandrR [38]3 years ago
8 0

Answer:16???

Step-by-step explanation:

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Rewrite the product as a sum: 10cos(5x)sin(10x)
mote1985 [20]

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)]

8 0
3 years ago
Willam fills 1/3 of a water bottle in 2/5 of a minute. How much time will it take him to fill the bottle?​
True [87]

Answer:

1/2 of a minute

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Hello people again please help
densk [106]
Ok so a triangle is equal to 180 degrees!!!
So with that info u can make a crazy little equation
180=70+68+47+x
Ok then you add 70,68 and 47
180=185+x
Ok now u have to subtract 185 on both sides
And get the answer -5=x or x=-5 (same thing)
3 0
2 years ago
The size of the interior angle of a regular decagon...Need an answer ASAP
olga nikolaevna [1]

Answer:

144

Step-by-step explanation:

x = ((n-2)π / n) radians = (((n-2)/n) x 180° ) degrees

5 0
3 years ago
Read 2 more answers
A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par
lyudmila [28]

Answer:

a(\frac{1}{5e})=5e

Step-by-step explanation:

we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

s'(t)=\ln \left(5t\right)+1

now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)

=\frac{1}{t}+0

a(t)=\frac{1}{t}

Firstly, we will set velocity =0

and then we can solve for t

v(t)=s'(t)=\ln \left(5t\right)+1=0

we get

t=\frac{1}{5e}

now, we can plug that into acceleration

and we get

a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}

a(\frac{1}{5e})=5e


5 0
3 years ago
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