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Leni [432]
3 years ago
6

5y-y^3=12 in verbal sentence

Mathematics
2 answers:
ExtremeBDS [4]3 years ago
4 0
Five y minus y to the power of 3 equals twelve
iogann1982 [59]3 years ago
4 0

Answer:

The difference of the 5 times a number and the number cubed is equal to 12.

Step-by-step explanation:

Consider the provided equation.

5y-y^3=12

Verbal expression: It is the expression which interprets a mathematical expression of operators variables into words.

Let y is the number,

5y can be read as 5 times of a number.

y³ can be read as the cube of the number.

We use the sign "-" for indicating the difference, the sign "=" is know as the equal to sign.

Now the verbal sentence for the equation 5y-y^3=12 is:

The difference of the 5 times a number and the number cubed is equal to 12.

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A machine wheel spins at a rate of 500 revolutions per minute. If the wheel has a diameter of 80 centimeters, what is the angula
solmaris [256]

Answer: 3,140 radians per minute.

Step-by-step explanation:

We know that the wheel does 500 revolutions per minute.

This is called the frequency of the wheel, and this is written as:

f = 500 rev/min = 500 RPM

The angular speed (or Angular velocity) is written as

ω = 2*pi*f

And this quantity is in radians/unit of time.

where pi = 3.14

then:

ω = 2*3.14*500 (rev/min)*(rad/rev) = 3,140 rad/min

This means that the angular velocity  (or angular speed) is 3,140 radians per minute.

8 0
3 years ago
Read 2 more answers
Consider the differential equationy′′+3y′−10y=0.(a) Find the general solution to this differential equation.(b) Now solve the in
RSB [31]

Answer:

Y= 2e^(5t)

Step-by-step explanation:

Taking Laplace of the given differential equation:

s^2+3s-10=0

s^2+5s-2s-10=0

s(s+5)-2(s+5) =0

(s-2) (s+5) =0

s=2, s=-5

Hence, the general solution will be:

Y=Ae^(-2t)+ Be^(5t)………………………………(D)

Put t = 0 in equation (D)

Y (0) =A+B

2 =A+B……………………………………… (i)

Now take derivative of (D) with respect to "t", we get:

Y=-2Ae^(-2t)+5Be^(5t)   ....................... (E)

Put t = 0 in equation (E) we get:

Y’ (0) = -2A+5B

10  = -2A+5B ……………………………………(ii)

2(i) + (ii) =>

2A+2B=4 .....................(iii)

-2A+5B=10 .................(iv)

Solving (iii) and (iv)

7B=14

B=2

Now put B=2 in (i)

A=2-2

A=0

By putting the values of A and B in equation (D)

Y= 2e^(5t)

6 0
4 years ago
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