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Masteriza [31]
3 years ago
5

What number does this Roman numeral represent? DCCLXXXI

Mathematics
2 answers:
avanturin [10]3 years ago
7 0

k there is your answer. be happy. UwU

Orlov [11]3 years ago
4 0
781 is DCCLXXXI in standard numbers
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What is the sum of (2x4 + 5x3 - 8x2 - x + 10) and (8x4 - 4x3 + x2 - x + 2)?
balu736 [363]

Option D: 10x^4+x^3-7x^2-2x+12 is the right answer

Step-by-step explanation:

Given polynomials are:

P_1 : 2x^4+5x^3-8x^2-x+10\\P_2: 8x^4-4x^3+x^2-x+2

We have to find the sum of polynomials

So,

S = P_1+P_2\\= (2x^4+5x^3-8x^2-x+10) + (8x^4-4x^3+x^2-x+2)\\=2x^4+5x^3-8x^2-x+10+8x^4-4x^3+x^2-x+2\\

Combining like terms

=2x^2+8x^4+5x^2-4x^3-8x^2+x^2-x-x+10+2\\= 10x^4+x^3-7x^2-2x+12

Hence,

Option D: 10x^4+x^3-7x^2-2x+12 is the right answer

Keywords: Polynomials, sum

Learn more about polynomials at:

  • brainly.com/question/2116906
  • brainly.com/question/2131336

#LearnwithBrainly

3 0
3 years ago
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The cost of a t-shirt is 8.50. David has $50 with him. How many t-shirt can he buy?​
anzhelika [568]
50÷8.5 = 5.88 but since we can’t have .88 of a person the answer would be five T-shirts
3 0
2 years ago
Original function is f(x)=2^x
Xelga [282]

Answer:

Look at the photo below for the answer.

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6 0
3 years ago
Let f(x)=x2+2x+3 . What is the average rate of change for the quadratic function from x=−2 to x = 5? Enter your answer in the bo
Crazy boy [7]

Answer:

Hello I just took the test and the answer is 5!

5 0
3 years ago
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For the following pair of functions, find (f+g)(x) and (f-g)(x).
ch4aika [34]

Given:

The functions are

f(x)=4x^2+7x-5

g(x)=-9x^2+4x-13

To find:

The functions (f+g)(x) and (f-g)(x).

Solution:

We know that,

(f+g)(x)=f(x)+g(x)

(f+g)(x)=4x^2+7x-5-9x^2+4x-13

(f+g)(x)=(4x^2-9x^2)+(7x+4x)+(-5-13)

(f+g)(x)=-5x^2+11x-18

And,

(f-g)(x)=f(x)-g(x)

(f-g)(x)=(4x^2+7x-5)-(-9x^2+4x-13)

(f+g)(x)=4x^2+7x-5+9x^2-4x+13

(f+g)(x)=(4x^2+9x^2)+(7x-4x)+(-5+13)

(f-g)(x)=13x^2+3x+8

Therefore, the required functions are (f+g)(x)=-5x^2+11x-18

and (f-g)(x)=13x^2+3x+8.

7 0
3 years ago
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