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3 years ago

PLEASE HELP WITH INEQUALITIES

Mathematics

2 answers:

lara31 [8.8K]3 years ago

8 0

Answer: 1 is greater 2 is less 3 is greater 4 is less

Step-by-step explanation:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

28. x > 1 (greater than)

29. x </= -1 (less than or equal to)

30. x >/= 3 (greater than or equal to)

31. x < 0 (less than)

Step-by-step explanation:

If the circle on the graph is open, it's "less than" or "greater than", but not "equal to". The open circle means that number is excluded from the answer set.

If the circle on the graph is closed, it's "less than or equal to" or "greater than or equal to"

Then the "less than" or "greater than" part depends on which way your arrow is pointing.

Please lmk if you have questions.

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Delia worked 45 minutes on her oil painting.She took a break at 10:35 a.m.At what time did Delia start working on her painting

rjkz [21]

Um i think it would be 9:55

4 0

3 years ago

Read 2 more answers

Which of these statements is true for f(x)=(1/10)^x

lana66690 [7]

Step-by-step explanation:

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

Analyzing option A)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

Putting $x = 1$ in the function

$f\left(1\right)=\:\left(\frac{1}{10}\right)^1$

$f\left(1\right)=\:\left\frac{1}{10}\right$

So, it is TRUE that when $x = 1$ then the out put will be $f\left(1\right)=\:\left\frac{1}{10}\right$

Therefore, the statement that '' The graph contains $\left(1,\:\frac{1}{10}\right)$ '' is TRUE.

Analyzing option B)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

The range of the function is the set of values of the dependent variable for which a function is defined.

$\mathrm{The\:range\:of\:an\:exponential\:function\:of\:the\:form}\:c\cdot \:n^{ax+b}+k\:\mathrm{is}\:\:f\left(x\right)>k$

$k=0$

$f\left(x\right)>0$

Thus,

$\mathrm{Range\:of\:}\left(\frac{1}{10}\right)^x:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)>0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(0,\:\infty \:\right)\end{bmatrix}$

Therefore, the statement that ''The range of $f(x)$ is $y > \frac{1}{10}$ " is FALSE

Analyzing option C)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

The domain of the function is the set of input values which the function is real and defined.

As the function has no undefined points nor domain constraints.

So, the domain is $-\infty \:$

Thus,

$\mathrm{Domain\:of\:}\:\left(\frac{1}{10}\right)^x\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

Therefore, the statement that ''The domain of $f(x)$ is $x>0$ '' is FALSE.

Analyzing option D)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

As the base of the exponential function is less then 1.

i.e. 0 < b < 1

Thus, the function is decreasing

Also check the graph of the function below, which shows that the function is decreasing.

Therefore, the statement '' It is always increasing '' is FALSE.

Keywords: function, exponential function, increasing function, decreasing function, domain, range

Learn more about exponential function from brainly.com/question/13657083

#learnwithBrainly

3 0

3 years ago

Read 2 more answers

PLEASE HELP<br><br> Graph the ordered pairs for y = 3x + 3 using r = {-2, 1, 2}.

Fed [463]

9514 1404 393

Answer:

(-2, -3), (1, 6), (2, 9) are plotted in the attached graph

Step-by-step explanation:

For x = -2, y = 3(-2) +3 = -3. The ordered pair is (-2, -3).

For x = 1, y = 3(1) +3 = 6. The ordered pair is (1, 6).

For x = 2, y = 3(2) +3 = 9. The ordered pair is (2, 9).

The graph is attached.

5 0

3 years ago

Read 2 more answers

Do fractions n always need to be renamed to the largest unit?

BabaBlast [244]

No, but if it's an improper fraction it should be done

5 0

3 years ago

Questions 16-17 | Math 1 - 0 points Solve the graph Help needed !!

iris [78.8K]

Answer:

16) The area of the circle is 25.1 units²

17) JKLM is a parallelogram but not a rectangle

Step-by-step explanation:

16) Lets talk about the area of the circle

- To find the area of the circle you must find the length of the radius

- In the problem you have the center of the circle and a point on

the circle, so you can find the length of the radius by using the

distance rule

* Lets solve the problem

∵ The center of the circle is (1 , 3)

∵ The point on the circle is (3 , 5)

- Using the rule of the distance between two points

* Lets revise it

- The distance between the two points (x1 , y1) and (x2 , y2) is:

Distance = √[(x2 - x1)² + (y2 - y1)²]

∴ r = √[(3 - 1)² + (5 - 3)²] = √[4 + 4] = √8 = 2√2 units

∵ The area of the circle = πr²

∴ The area of the circle = π (2√2)² = 8π = 25.1 units²

* The area of the circle is 25.1 units²

17) To prove a quadrilateral is a parallelogram, prove that every

to sides are parallel or equal or the two diagonal bisect

each other

* The parallelogram can be rectangle if two adjacent sides are

perpendicular to each other (measure of angle between them is 90°)

or its diagonals are equal in length

- The parallel lines have equal slopes, then to prove the

quadrilateral is a parallelogram, we will find the slopes of

each opposite sides

* Lets find from the graph the vertices of the quadrilateral

∵ J = (0 ,2) , K (2 , 5) , L (5 , 0) , M (3 , -3)

- The opposite sides are JK , ML and JM , KL

- The slope of any line passing through point (x1 , y1) and (x2 , y2) is

m = (y2 - y1)/(x2 - x1)

∵ The slop of JK = (5 - 2)/(2 - 0) = 3/2 ⇒ (1)

∵ The slope of LM = (-3 - 0)/(3 - 5)= -3/-2 = 3/2 ⇒ (2)

- From (1) and (2)

∴ JK // LM

∵ The slope of KL = (0 - 5)/(5 - 2) = -5/3 ⇒ (3)

∵ The slope of JM = (-3 - 2)/(3 - 0)= -5/3 ⇒ (4)

- From (3) and (4)

∴ KL // JM

∵ Each two opposite sides are parallel in the quadrilateral JKLM

∴ It is a parallelogram

- The product of the slopes of the perpendicular line is -1

* lets check the slopes of two adjacent sides in the JKLM

∵ The slope of JK = 3/2 and the slope of KL = -5/3

∵ 3/2 × -5/3 = -5/2 ≠ -1

∴ JKLM is a parallelogram but not a rectangle

4 0

3 years ago