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Flauer [41]
3 years ago
5

How to solve for x on a square root

Mathematics
1 answer:
enot [183]3 years ago
3 0
You would take the answer of the square root problem and find the number that you can multiply by its self that would make that answer correct.
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What is the domain of the function f(x) = x + 2 ? a.    all real numbers greater         than -2 b.    all real numbers greater
dsp73
The domain of the function is represented by option C. All real numbers.
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3 years ago
What algebraic expression must be added to the sum of 3x2 +4x+8 and 2x2−6x+3 to give 9x2−2x−5 as the result?
vesna_86 [32]

Answer:

4x^2-16

Step-by-step explanation:

-This is an addition/subtraction problem.

#We subtract the sum of the first two expression from the third expression:

\#Add \ first \ two\\\\=3x^2 +4x+8 +2x^2-6x+3\\\\=5x^2-2x+11\\\\\# Subtract\ from \ the \ third\\=(9x^2-2x-5)-(5x^2-2x+11)\\\\=9x^2-2x-5-5x^2+2x-11\\\\=4x^2-16

Hence, the expression 4x^2-16 must be added to the first two.

7 0
3 years ago
Question 31 pts Prove the statement is true using mathematical induction: 2n-1 ≤ n! Use the space below to write your answer. To
Sladkaya [172]

Answer:

P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Step-by-step explanation:

Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3

Basis: P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:

2(k+1) - 1 = 2k + 2 - 1

≤ 2 + k! (by the inductive hypothesis)

= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.

4 0
3 years ago
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natulia [17]

Answer:

19

Step-by-step explanation:

The pattern is that it +6 every number.

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1 + 6 = 7

7 + 6 = 13

So the next number is 13 + 6 = 19.

EDIT - I can't add sorry.

4 0
3 years ago
Read 2 more answers
Will mark as brainliest!!
fredd [130]
14 feet I think..........................
3 0
3 years ago
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