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zimovet [89]
3 years ago
12

Consider the following polynomial.

Mathematics
2 answers:
LenaWriter [7]3 years ago
4 0
Answers:
1) 6
2) 5
3) 8
4)5ab^5
Mkey [24]3 years ago
4 0

-2ab^3b^2+5ab^5+7b^4+8

As we can see that the exponents over the variables of first term are 1 , 3 and 2 , so total becomes 1+3+2 = 6 ,

Exponents over variables of second term are 5

Exponent over variables of third term is 4 and

Exponent over the variables of last term is 0 .

It means they are already arranged in the proper form of a polynomial.

-2ab^3b^2+5ab^5+7b^4+8


Hence

a) Degree = 1+3+2 = 6

b) Leading coefficient = -2

c) Constant term = 8

d) leading term = -2ab^3b^2

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If m 2 n, which inequalities must be true? Check all that apply.
anzhelika [568]

Answer:

(a)\ m+ 2.1 > n + 2.1

(b).\ m - (-4) > n - (-4)

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Step-by-step explanation:

Given

m > n

Required

Select true inequalities

(a)\ m+ 2.1 > n + 2.1

Subtract 2.1 from both sides

m > n

This is true

(b).\ m - (-4) > n - (-4)

Open brackets

m + 4 > n + 4

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3 years ago
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4 years ago
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

Urn\ 1 = \{B_1, R_1, R_2, R_3\}

Urn\ 2 = \{R_4, R_5, B_2, B_3\}

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

<em>See attachment for possibility tree</em>

Solving (b): The total number of outcome

<u>For urn 1</u>

There are 4 balls in urn 1

n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

Urn\ 1 = 12

<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

So, the selection is:

Urn\ 2 = 4 * 3

Urn\ 2 = 12

Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

Total = 24

5 0
3 years ago
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