2 years ago

Consider the piecewise function shown on the graph, which is composed of polynomial, constant, and exponential pieces.

Mathematics

1 answer:

BabaBlast [244]2 years ago

4 0

The correct answer is D.

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Need help please, also about probabilities

liberstina [14]

Answer:

Step-by-step explanation:

Good evening ,

_______________

a) p(rat)=1-(0,35+0,4+0,1)

b) p(cat ∪ hamster)=0,35+0,1

c) p(dog then dog)= (0,4)²

8 0

3 years ago

TiliK225 [7]

What I’m so confused

3 0

3 years ago

Read 2 more answers

The vertices of APQR are P(-3, 8) Q(-6,-4), and R(1, 1). If you reflect APQR across the x-axis, what will be the coordinates of

Ymorist [56]

Answer:(-3,-8),(-6,4),(1,-1)

Step-by-step explanation:

Given

Vertices of the triangle are

$P(-3,8), Q(-6,-4),R(1,1)$

Coordinates (a,b) when reflected about x-axis is given as (a,-b) i.e. Y coordinate changes its sign

$\therefore P\rightarrow P'(-3,-8)\\\Rightarrow Q\rightarrow Q'(-6,4)\\\Rightarrow R\rightarrow R'(1,-1)$

4 0

2 years ago

What is (-4) raised to the negative third power?

Usimov [2.4K]

-4(-4) = positive 16
Positive 16(-4) = -64
-64 is your answer

7 0

3 years ago

Read 2 more answers

Select two polynomials that have a sum of 16n^3+13n^2+7n-10 A. 8n^3 - 5n +11n^2 - 5 B. 7n^2 + 8n - n^3 + 2 C. 15n^2 - 10n + 3n^3

andreyandreev [35.5K]

Answer:

Step-by-step explanation:

We have the polynomial

$16n^3+13n^2+7n-10$ ( 1 )

To solve this problem we have to take into account that only we can sum term with the same order in the variable. We have the polynomials

$8n_{3}-5n+11n^{2}-5\\7n^2 + 8n - n^3 + 2\\15n^2 - 10n + 3n^3 - 4\\2n^2 + 12n - 5 + 8n^3$

we can note (by summing term by term) that only the sum of the first and the fourth equation correspond to the given polynomial ( 1 ) of the problem. If we organize these polynomials (that is, write the equation down in a form where higher order appears first ) we have

$8n^3 +11n^2 - 5n - 5\\ 8n^3+2n^2 + 12n - 5\\$

and if we sum we obtain

$16n^3+13n^2+7n-10$

that is what we was looking for

I hope this is useful for you

regards

7 0

2 years ago