All categories

3 years ago

For the piecewise Function below, which of the following statments is true? Someone please help me

Mathematics

2 answers:

kirill115 [55]3 years ago

8 0

Remark

Let's just figure out what f(1) and f(3) are and what equation to use.

f(1)

f(1) is defined by the top condition. Look at the condition carefully. -3 ≤ x ≤ 1

So the value of f(1) is

(x + 1)² -1

(1 + 1)² - 1

2² - 1

4 - 1 = 3

f(3)

f(3) is governed by the middle condition. 1 < x ≤ 3

The value of f(3) is - x + 2 = - 3 + 2 = - 1

Conclusion

f(1) > f(3)

Comment

Both f(1) and f(3) are defined. You just have to look around to find out where. D is wrong.

We calculated both f(1) and f(3). f(1) is larger than f(3) so that is your answer.

AysviL [449]3 years ago

4 0

The first one and the third one are correct. because for the when you plug 1 on the first piece wise function you get 3.

f(3) =1. and f(1)=3 so f(1)>f(3)

You might be interested in

PLEASE HELP!!!! MANY POINT GIVEN

NikAS [45]

Answer: 36

Step-by-step explanation:

11 x 2= 22 ÷ 2= 11

10 x 5= 50÷ 2= 25

11+25= 36

5 0

3 years ago

Read 2 more answers

Triangle ABCis rotated 90° clockwise around the origin to get

Elanso [62]

Answer:

The distance A’C’ is 4.47 units

Step-by-step explanation:

Before we go on, we need to get the appropriate transformation

Mathematically, we have a 90 degrees clockwise rotation yielding the following;

(x,y) to (-y,x)

A is (-4,1)

C is (-2,5)

By transforming, we have

A’( -1,-4)

C’ (-5,-2)

To get the magnitude of the line segment, we are going to use the distance formula between points

We have this as;

D = √(x2-x1)^2 + (y2-y1)^2

D = √(-5-(-1))^2 + (-2-(-4))^2

D = √(-4)^2 + (2)^2

D = √(16 + 4)

D = √20

D = 4.47 units

7 0

3 years ago

The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet

vladimir1956 [14]

$\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}$

$\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
-------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}$

8 0

3 years ago

Read 2 more answers

Alicia has invented a new app that two companies are interested in purchasing for a 2-year contract.

9966 [12]

8th payment, if you want how to get the answer I made a chart as shown in the picture

8 0

3 years ago

Read 2 more answers

A rectangular sheet of paper was folded in half 6 times. In the middle of this folded sheet, two holes were drilled. Then, the s

laiz [17]

Answer:

There will be 128 holes

Step-by-step explanation:

Simply think of each fold as doubling the number of holes. So since we have 6 folds, that will be 2^6 and then we have 2 holes in those folds, which makes 2*2^6 == 2^7 == 128 holes. Cheers.

7 0

4 years ago