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Phantasy [73]
1 year ago
7

natural numbers will ____ be whole numbers. Integers will ___ be natural numbers. Irrational numbers will ___ be real numbers. R

ational numbers will be irrational numbers. Rational numbers will __ integers..Fill in the blank in each sentence with either "Always", "sometimes", "never"
Mathematics
1 answer:
NARA [144]1 year ago
3 0

- Natural numbers will Always be whole numbers.

Natural numbers are positive integers (whole numbers), so, they are always whole numbers.

- Integers will Sometimes be natural numbers.

Not all integers are natural numbers. Negative integers are not natural numbers while positive integers are natural numbers.

- Irrational numbers will Always be real numbers.

Real numbers consist of both irrational and rational numbers. So, all irrational numbers are real numbers.

- Rational numbers will Never be irrational numbers.

Rational numbers and irrational numbers are two different types of real numbers.

- Rational numbers will Sometimes be integers.

Not all rational numbers are integers. there are some rational numbers that are not integers while some rational numbers are integers.

er

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Given g(x)=-3x-4 what is the x value when g(x)=2
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Answer:

x= -2

Step-by-step explanation:

2= -3(x)-4

6= -3x

-2= x

7 0
3 years ago
Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)
DiKsa [7]

Answer:

C

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. y'=(3\sin x)'=3\cos x. Differentiate again to get y''=(3\cos x)'=-3\sin x, then y''+y=-3\sin x+3\sin x=0\neq 3\sin x so this choice of y doesn't solve the equation.

B. y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x and y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x, then y''+y=10\sin x+6\cos x\neq 3\sin x so y is not a solution

C. y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x hence y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x. Then y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x so y is a solution.

D.y'=-3\sin x and y''=-3\cos x, then y''+y=0 thus y isn't a solution

E. y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x hence y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x. Then y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x then y is not a solution.

3 0
3 years ago
Please answer a and b
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Answer:

(-2,1) (-1,1),(0,1)

Step-by-step explanation:

find the gradient first

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murzikaleks [220]

Step-by-step explanation:

we know that

Work =f*d

=25*100

=2500Joule

therefore work done is 2500Joule

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Can anyone solve this ?
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Is e just an algebraic expression or is it Euler's number (2.718...)?

If e is an algebraic expression then there it is

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