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3 years ago

How would you solve cos(5x)=sin(10x) ?

1 answer:

4 0

We will use double angle identities:
cos (5x ) = sin (10x )
cos (5x ) = 2 cos (5x ) sin ( 5x )
cos ( 5 x) - 2 cos ( 5 x ) sin ( 5x ) = 0
cos ( 5 x ) · [ 1 - 2 sin (5 x) ] = 0
cos ( 5 x ) = 0 or :   1 - 2 sin (5 x) = 0
5 x = π/2 +kπ, k∈Z sin (5 x) = 1/2
x1 = π/10 + kπ/5    5 x = π/6+2kπ , k∈ Z
5 x = 5π/6 +2kπ , k∈ Z
x 2 = π/30 +2kπ/5
x 3 = π/9 + 2kπ/5

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Solve 7x/3< 2. please

antiseptic1488 [7]

Answer:

x < 6/7

Step-by-step explanation:

Multiply both sides by 3 to get 7x < 6

Divide by 7

x < 6/7

6 0

4 years ago

HELP! HELP! HELP! A function and its inverse are shown on the graph. Which answer pairs a possible domain restriction for f(x) a

Svetradugi [14.3K]

Answer:

B. f(x) domain: x ≥ 1; f⁻¹(x) range: y ≥ 1

Step-by-step explanation:

The <em>domain</em> of a function is identical to the <em>range</em> of its inverse. This is reflected in choices B and D. However, f(x) is undefined for x < 1, so it makes no sense to restrict its domain to x ≤ -2, as in choice D.

The appropriate response is ...

f(x) domain: x ≥ 1
f⁻¹(x) range: y ≥ 1

6 0

3 years ago

Read 2 more answers

a submarine ascended 3/8 mile and the descended .4 of a mile followed by another 2/5 mile descent, finally it rose .75 of a mile

shusha [124]

Do you still need the answer? lol.

5 0

4 years ago

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

If the product of two numbers is equal to zero then at least one of the numbers must be zero

Zarrin [17]

Answer:

True

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers