Find f(–3) for the following function

A set X consists of all real numbers greater than or equal to 1. Use set-builder notation to define X.

vitfil [10]

Theory:

The standard form of set-builder notation is 

{ x | “x satisfies a condition” }

This set-builder notation can be read as “the set of all x such that x (satisfies the condition)”.

For example, { x | x > 0 } is equivalent to “the set of all x such that x is greater than 0”.

Solution:

In the problem, there are 2 conditions that must be satisfied:

1st: x must be a real number

In the notation, this is written as “x ε R”. Where ε means that x is “a member of” and R means “Real number”

2nd: x is greater than or equal to 1

This is written as “x ≥ 1”

Answer:

Combining the 2 conditions into the set-builder notation:

X = { x | x ε R and x ≥ 1 }

8 0

3 years ago

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How to do this? Angles ?

gayaneshka [121]

Answer:

this is beyond me I'm jus try complete my verification

Step-by-step explanation:

sorry but good luck

6 0

3 years ago

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Brainly sound waves travel at about 1000 feet per second. Loght travels at 9.8×10^8 feet per second. How many times faster does

Fantom [35]

Answer:

980,000 times faster

Step-by-step explanation:

Please kindly see the attached file for more explanation

3 0

4 years ago

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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 2 (a) Find the interval on which

Burka [1]

Answer:

a) $(-\infty, -1) \cup (5, \infty)$

b) $(-1,5)$

Step-by-step explanation:

The first step to solve this question is finding the roots of the derivative of x.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$f(x) = x^{3} - 6x^{2} - 15x + 2$

So

$f'(x) = 3x^{2} - 12x - 15$

Finding the roots:

$3x^{2} - 12x - 15 = 0$

Simplifying by -3

$x^{2} - 4x - 5 = 0$

So $a = 1, b = -4, c = -5$

Then

$\bigtriangleup = (-4)^{2} - 4*1*(-5) = 36$

$x_{1} = \frac{-(-4) + \sqrt{36}}{2} = 5$

$x_{2} = \frac{-(-4) - \sqrt{36}}{2} = -1$

So the function can be divided in three intervals.

They are:

Less than -1

Between -1 and 5

Higher than 5

In which it increases and which it decreases?

Less than -1

Lets find the derivative in a point in this interval, for example, -2

$f'(x) = 3x^{2} - 12x - 15$

$f'(-2) = 3*(-2)^{2} - 12*(-2) - 15 = 21$

Positive.

So in the interval of $(-\infty, -1)$ , the function increases.

Between -1 and 5

Will choose 0.

$f'(x) = 3x^{2} - 12x - 15$

$f'(0) = 3*(0)^{2} - 12*(0) - 15 = -15$

Negative.

So in the interval of $(-1,5)$ , the function decreases.

Higher than 5

Will choose 6.

$f'(x) = 3x^{2} - 12x - 15$

$f'(6) = 3*(6)^{2} - 12*(6) - 15 = 21$

Positive

So in the interval of $(5, \infty)$ , the function increases.

(a) Find the interval on which f is increasing.

Using interval notation

$(-\infty, -1) \cup (5, \infty)$

b) Find the interval on which f is decreasing.

$(-1,5)$

5 0

3 years ago

Find the fifth term of the arithmetic sequence in which t1 = 3 and tn = tn-1 + 4. A) 5 B) 7 C) 19 D) 23 2) Find the tenth term o

Sergeu [11.5K]

We know that

Part a) Find the fifth term of the arithmetic sequence in which t1 = 3 and tn = tn-1 + 4
t1=3
t2=t1+4----> 3+4-----> 7
t3=t2+4-----> 7+4----> 11
t4=t3+4-----> 11+4---> 15
t5=t4+4-----> 15+4---> 19

the answer Part a) is 19

Part b) Find the tenth term of the arithmetic sequence in which t1 = 2 and t4 = -10

we know that
tn=t1+(n-1)*d-----> d=[tn-t1]/(n-1)
t1=2
t4=-10
n=4
find the value of d
d=[-10-2]/(4-1)-----> d=-12/3----> d=-4
find the tenth term (t10)
t10=t1+(10-1)*(-4)----> t10=2+9*(-4)----> t10=-34

the answer Part b) is -34

Part c) Find the fifth term of the geometric sequence in which t1 = 3 and tn = 2tn-1
t1=3
t2=2*t1----> 2*3----> 6
t3=2*t2----> 2*6----> 12
t4=2*t3-----> 2*12---> 24
t2=2*t4----> 2*24----> 48

the answer Part c) is 48

3 0

4 years ago