I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Which equation represents a line through (3,5) that is perpendicular to y=2x-5?

tiny-mole [99]

I think it 3 bit I am not sure

7 0

4 years ago

danny has 45 minutes to take the math test if danny finishes the test in 19 minutes how many minutes does he have left to the fi

____ [38]

Just take 45 minutes minus 19 and you will get 26 minutes left

6 0

3 years ago

Read 2 more answers

What is this answer I need help 3x^2-48=0

devlian [24]

Answer: the answer is X=8

Step-by-step explanation:

3 0

3 years ago

1 х Given g(x)=

yarga [219]

(a) Since $g(x)=\sqrt[3]{x}$ and $h(x) = \frac1{x^3}$ , we have

$(g\circ h)(x) = g(h(x)) = g\left(\dfrac1{x^3}\right) = \sqrt{3}{\dfrac1{x^3}} = \dfrac1x$

We're given that

$(f \circ g \circ h)(x) = f(g(h(x))) = f\left(\dfrac1x\right) = \dfrac x{x+1}$

but we can rewrite this as

$\dfrac x{x+1} = \dfrac{\frac xx}{\frac xx + \frac1x} = \dfrac1{1+\frac1x}$

(bear in mind that we can only do this so long as x ≠ 0) so it follows that

$f\left(\dfrac1x\right) = \dfrac1{1+\frac1x} \implies \boxed{f(x) = \dfrac1{1+x}}$

(b) On its own, we may be tempted to conclude that the domain of $(f\circ g\circ h)(x) = \frac1{1+x}$ is simply x ≠ -1. But we should be more careful. The domain of a composite depends on each of the component functions involved.

$g(x) = \sqrt[3]{x}$ is defined for all x - no issue here.

$h(x) = \frac1{x^3}$ is defined for all x ≠ 0. Then $(g\circ h)(x) = \frac1x$ also has a domain of x ≠ 0.

$f(x) = \frac1{1+x}$ is defined for all x ≠ -1, but

$(f\circ g\circ h)(x)=f\left(\frac1x\right) = \dfrac1{1+\frac1x}$

is undefined not only at x = -1, but also at x = 0. So the domain of $(f\circ g\circ h)(x)$ is

$\left\{x\in\mathbb R \mid x\neq-1 \text{ and }x\neq0\right\}$

7 0

3 years ago

An urn contains two blue balls (denoted B1 and B2) and three white balls (denoted W1, W2, and W3). One ball is drawn, its color

alekssr [168]

Answer:

(a) Shown below.

(b) The probability that the first ball drawn is blue is 0.40.

(c) The probability that only white balls are drawn is 0.36.

Step-by-step explanation:

The balls in the urn are as follows:

Blue balls: B₁ and B₂

White balls: W₁, W₂ and W₃

It is provided that two balls are drawn from the urn, with replacement, and their color is recorded.

(a)

The possible outcomes of selecting two balls are as follows:

B₁B₁ B₂B₁ W₁B₁ W₂B₁ W₃B₁

B₁B₂ B₂B₂ W₁B₂ W₂B₂ W₃B₂

B₁W₁ B₂W₁ W₁W₁ W₂W₁ W₃W₁

B₁W₂ B₂W₂ W₁W₂ W₂W₂ W₃W₂

B₁W₃ B₂W₃ W₁W₃ W₂W₃ W₃W₃

There are a total of N = 25 possible outcomes.

(b)

The sample space for selecting a blue ball first is:

S = {B₁B₁, B₁B₂, B₁W₁, B₁W₂, B₁W₃, B₂B₁, B₂B₂, B₂W₁, B₂W₂, B₂W₃}

n (S) = 10

Compute the probability that the first ball drawn is blue as follows:

$P(\text{First ball is Blue})=\frac{n(S)}{N}=\frac{10}{25}=0.40$

Thus, the probability that the first ball drawn is blue is 0.40.

(c)

The sample space for selecting only white balls is:

X = {W₁W₁, W₂W₁, W₃W₁, W₁W₂, W₂W₂, W₃W₂, W₁W₃, W₂W₃, W₃W₃}

n (X) = 9

Compute the probability that only white balls are drawn as follows:

$P(\text{Only White balls})=\frac{n(X)}{N}=\frac{9}{25}=0.36$

Thus, the probability that only white balls are drawn is 0.36.

4 0

3 years ago