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The United States is not a direct democracy, in the sense of a country in which laws (and other government decisions) are made predominantly by majority vote.

Explanation:

Some lawmaking is done this way, on the state and local levels, but it’s only a tiny fraction of all lawmaking. But we are a representative democracy, which is a form of democracy.

The United States is also a constitutional democracy, in which courts restrain in some measure the democratic will. And the United States is therefore also a constitutional republic.

3 0

2 years ago

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Determine whether or not the vector field is conservative. If it is conservative, find a function f such that f = ∇f. (if the ve

Arisa [49]

If F(x, y, z) = i + sin(z) j + y cos(z) k is conservative, then there exists a scalar function f(x, y, z) such that grad(f) = F, which means

∂f/∂x = 1

∂f/∂y = sin(z)

∂f/∂z = y cos(z)

Integrating each each of these equations gives

∫ ∂f/∂x dx = ∫ dx ⇒ f(x, y, z) = x + α(y, z)

∫ ∂f/∂y dy = ∫ sin(z) dy ⇒ f(x, y, z) = y sin(z) + β(x, z)

∫ ∂f/∂z dx = ∫ y cos(z) dz ⇒ f(x, y, z) = y sin(z) + γ(x, y)

It follows that α(y, z) = y sin(z) and β(x, z) + γ(x, y) = x + C where C is a constant. So

f(x, y, z) = x + y sin(z) + C

and F is indeed conservative.

6 0

2 years ago

For how many three digit numbers (100 to 999) is the sum of the digits even?.

attashe74 [19]

The digital sum of a 3-digit number is even if

• all 3 of its digits are even, or

• 2 of its digits are odd and 1 is even

Count the number of 3-digit numbers, all of whose digits are even:

• the first digit cannot be 0, and otherwise has 4 options (2, 4, 6, or 8)

• the second and third digits each have 5 options (2, 4, 6, 8, or 0)

So there are 4 × 5² = 100 such numbers.

Count the number of 3-digit numbers with 2 odd digits and 1 even digit:

• if the first digit is even, then there are 4 options (2, 4, 6, or 8), and the other two digits each have 5 options (1, 3, 5, 7, or 9)

• if the first digit is odd, then there are 5 options (1, 3, 5, 7, or 9); if the second digit is also odd, it has the same 5 options; then the third digit is even and has 5 options (2, 4, 6, 8, or 0)

Swapping the second and third digits in the last case gets the same amount of numbers; we're effectively multiplying by the number of permutations of 2 objects taken 2 at a time, 2!.

So there are 4 × 5² + 2! × 5³ = 350.

Then there is a total of 450 3-digit numbers whose digital sums are even.

7 0

2 years ago