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3 years ago

What is the domain of f(x) equals 3X -2

Mathematics

1 answer:

Irina18 [472]3 years ago

3 0

Answer:

(0,7/3]

I think this is correct, not sure though. Hoping this still hepled a bit!!

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A person flips a coin 26 times and observes 41% of times head. Then, based on her observation, she claims that the probability o

Step2247 [10]

Answer: -Empirical probability-

Step-by-step explanation:

Because experimental probability of an event is the ratio of the number of outcomes in which a specified event occurs to the total number of trials.

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3 years ago

Three relationships are described below: I. The amount of fuel used on a trip increases as the size of the car increases and as

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Variation are of 2 types: Direct variation that states 'if x increases, y increases'. And inverse variation which states ' if x increases, y decreases and y increases x increases'.

1. The amount of fuel used on a trip increases as the size of the car increases and as the distance traveled increases. = Direct variation

2. As the number of people helping mow a lawn increases, the time it takes to mow the lawn decreases = Inverse variation

3. The cost of having a house painted increases as the size of the house increases. = Direct variation

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3 years ago

Hi! Okay I have a review from the 5th grade if someone can help me because I don't remember fractions it would be so helpful bec

DIA [1.3K]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find all the zeros of the equation. -x^3-3x^2=6x+4

k0ka [10]

Answer:

The zeros of given expression $-x^3-3x^2=6x+4$ is -1, $-1+i\sqrt{3}$ and $-1-i\sqrt{3}$

Step-by-step explanation:

Given expresssion is $-x^3-3x^2=6x+4$

To find zeros of given expression we have to equate the expression to zero.

ie., $-x^3-3^2-6x-4=0$

$-(x^3+3x^2+6x+4)=0$

$x^3+3x^2+6x+4=0$

By using synthetic division

-1 | 1 3 6 4

| 0 -1 -2 -4

|________________

1 2 4 0

Therefore (x+1) is a zero

Now the quadratic equation is $x^2+2x+4=0$

For quadratic equation $ax^2+bx+c=0$ we have

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here a=1 ,b=2 and c=4 now substitute the values

$x=\frac{-2\pm \sqrt{2^2-4(1)(4)}}{2(1)}$

$x=\frac{-2\pm \sqrt{4-16}}{2}$

$x=\frac{-2\pm \sqrt{-12}}{2}$

$x=\frac{-2\pm \sqrt{12i^2}}{2}$ where $i^2=-1$

$x=\frac{-2\pm i \sqrt{4\times 3}}{2}$

$x=\frac{-2\pm i \sqrt{4}\sqrt{3}}{2}$

$x=\frac{-2\pm 2i\sqrt{3}}{2}$

$x=2\times\frac{(-1\pm i\sqrt{3})}{2}$

$x=-1\pm i\sqrt{3}$

Therefore $x=-1+i\sqrt{3}$ and $x=-1-i\sqrt{3}$

Therefore the zeros are -1, $-1+i\sqrt{3}$ and $-1-i\sqrt{3}$

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4 years ago

Nelson is selling his photographs at an art festival the festival is open for 6 hours each day for 3 days at the conclusion of t

Lina20 [59]

Answer:

Step-by-step explanation:

$3*6=18$

$54/18=3$

Since he sold for 6 hours over the course of 3 days. Convert that to 18 hours total selling time. divide the amount sold by the hours and there is your answer.

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3 years ago