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Alja [10]
3 years ago
7

If like line A is changed to line B how will the equation change

Mathematics
1 answer:
Alla [95]3 years ago
3 0

Answer:

Line B will become line A and Line A will become line B.

Step-by-step explanation:

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The mean is defined as a
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All the numbers added up then divided by the amount of numbers.

EX: 5 + 10 + 3 + 8 + 9
       35
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3 years ago
If u = 8 + 4i and v = 3i + 6, then what is u – v?
lozanna [386]
u=8+4i\ and\ v=31+6\\\\u-v=(8+4i)-(3i+6)=8+4i-3i-6=(8-6)+(4i-3i)\\\\\huge\boxed{=2+i}\leftarrow \boxed{d}
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3 years ago
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
Use the following expression to
IgorLugansk [536]

Answer:

part A :BODMAS ( Bracket of Division Multiplication Addition Subtraction )

part b; Bracket (2+3.2) = - 2- 3.2

= 16- 2-3.2 +30-5

ADDITION:-3.2+30 = - 33.2

16-2(-33.2)30-5

14-33.2 +30-5

14-33.2 +25

-19.2+25

-44.2

Step-by-step explanation:

THE FOLLOWING EQUATION CAN ONLY BE SOLVED USING BODMAS

3 0
3 years ago
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