If the denominar is 25 what is the numrater

A line passes through the point (-8, 6) and has a slope of -5/4.

Margarita [4]

Answer:

y = -5/4x - 4

Step-by-step explanation:

Plug in 6 for y and -8 for x. Then solve for b or "y-intercept". You should get -4

5 0

2 years ago

Read 2 more answers

The notation f:S→T denotes that f is a function, also called a map , defined on all of a set S and whose outputs lie in a set T

AnnyKZ [126]

Answer:

There are many. Two examples are

$f(x) = x, \\f(x) = x^3$

Step-by-step explanation:

There are many examples. The simplest is

1 -

$f(x) = x$

It is trivial that

$\text{if \,\,\,\,} f(x) = f(y) \,\,\,\,\,\text{then} \,\,\,\,\, x=y$

2 -

$f(x) = x^3$

That function is injective as well.

$\text{if \,\,\,\,} x^3 = y^3 \,\,\,\,\,\text{then} \,\,\,\,\, x=y$

An example of a function that is NOT injective is

$f(x) = x^2$

Notice that

$f(-2) = (-2)^2 = 2^2 = 4$

4 0

3 years ago

4 thousands 7 hundreds = 47

motikmotik

If there is 4 thousands and 7 hundreds is 4,700

4 0

2 years ago

Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f

quester [9]

Answer:

It is continuous since $\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$

Step-by-step explanation:

We are given that the function is defined as follows $f(x) = 9-x, x\leq 0$ and $f(x) = 9+12x, x>0$ and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity) x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all $\mathbb{R}$ . So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

$\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$ (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

$\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9$ . This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

$\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9$ . As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, $\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9$ , so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

5 0

3 years ago

Read 2 more answers

In a jar, there is a green, a blue, and a red marble. You draw (with replacement) a marble from the jar until you get the green

denpristay [2]

Answer:

The value is $P( X = m ) = (\frac{2}{3} )^{m-1 } * \frac{1}{3}$

Step-by-step explanation:

From question we are told that

The number of types of marbles present jar is n = 3

Generally the probability of drawing a green marble is

$P(g) = \frac{1}{3}$

Generally the probability of drawing a marble that is not green is

$P(g') = 1 - \frac{1}{3} = \frac{2}{3}$

From the question we are told that there will be continuous drawing of marbles from the jar (in such a way that after each marble is drawn it is being replaced) until a green marble drawn

Let m be the number of times marbles has been drawn when a green marble was gotten

it then means that for m - 1 times the marbles where drawn a green marble was not obtain.

Generally the probability drawing m times is mathematically is mathematically represented as

$P( X = m ) = (P(g'))^{m-1 } * P(g)$

=> $P( X = m ) = (\frac{2}{3} )^{m-1 } * \frac{1}{3}$

7 0

2 years ago