A line with a slope of 1 passes through the point (0, 2)What is its equation in slope-intercept form?

Mathematics

1 answer:

devlian [24]3 years ago

5 0

Answer:

y=x+2

Step-by-step explanation:

y-y1=m(x-x1)

y-2=1(x-0)

y-2=1(x)

y-2=x

y=x+2

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What is the solution to ln (x2 - 16) = 0?

bogdanovich [222]

ANSWER

$x = \pm \sqrt{17}$
EXPLANATION

The given equation is

$ln( {x}^{2} - 16) = 0$

Take antilogarithm of both sides to base e.

${e}^{ ln( {x}^{2} - 16) } = {e}^{0}$

This will simplify to

${x}^{2} - 16 = 1$

Group like terms to obtain,

${x}^{2} = 1 + 16$

${x}^{2} = 17$

Take square root of both sides to get,

$x = \pm \sqrt{17}$

6 0

3 years ago

Read 2 more answers

Let n be a positive integer and define [n] to be the set of the first n positive integers. That is, [n] = {1, 2, 3, . . . , n}.

yaroslaw [1]

Answer: There are $2^{n-1}$ ways of doing this

Hi!

To solve this problem we can think in term of binary numbers. Let's start with an example:

n=5, A = {1, 2 ,3}, B = {4,5}

We can think of A as 11100, number 1 meaning "this element is in A" and number 0 meaning "this element is not in A"

And we can think of B as 00011.

Thinking like this, the empty set is 00000, and [n] =11111 (this is the case A=empty set, B=[n])

This representation is a 5 digit binary number. There are $2^5$ of these numbers. Each one of this is a possible selection of A and B. But there are repetitions: 11100 is the same selection as 00011. So we have to divide by two. The total number of ways of selecting A and B is the $2^{5-1} = 2^4$ .