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STatiana [176]
2 years ago
9

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
Previous Page
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
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