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Natali5045456 [20]
3 years ago
12

No links or trolls please. I need help with this math, it’s due soon. Thanks you. It’s a division sign by the way, it’s hard to

see in the photo.

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
3 0

Answer:

n^2p

Step-by-step explanation:

Thats the answer

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How is adding positive and negative fractions similar to adding intergers ?
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Answer:

When you are adding or subtracting a negative fraction, you usually want to consider the numerator as negative. The method is just the same, except now you may need to add negative or positive numerators. Example 1: ... To add the fractions with unlike denominators, rename the fractions with a common denominator.

Step-by-step explanation:

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3 years ago
What is the volumenof the sphere rounded to the nearest hunderths?​
Kryger [21]
9202.77 it should be
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3 years ago
Find the volume (Don't forget your UNITS) *
Radda [10]

Answer:

135 cm³

Step-by-step explanation:

  1. To find volume: you multiply length*width*height

    2. l*w*h

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5 0
3 years ago
Select all of the expressions that are
Gwar [14]

|6|  is equal to 6

|-6|  is equal to 6

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5 0
3 years ago
Read 2 more answers
What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!
Radda [10]

Two main facts are needed here:

1. The logarithm \log x, regardless of the base of the logarithm, exists for x>0.

2. The square root \sqrt x exists for x\ge0.

(in both cases we're assuming real-valued functions only)

By (2) we know that \sqrt{2x-1} exists if 2x-1\ge0, or x\ge\dfrac12.

By (1), we know that \log(\sqrt{2x-1}+3) exists if \sqrt{2x-1}+3>0, or \sqrt{2x-1}>-3. But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require x\ge\dfrac12, so the function has domain \left[\dfrac12,\infty).

To determine the range, we need to know that, in their respective domains, \sqrt x and \log x increase monotonically without bound. We also know that x=\dfrac12 at minimum, at which point the square root term vanishes, so the least value the function takes on is \log3. Then its range would be [\log3,\infty).

3 0
3 years ago
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