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Mashcka [7]
3 years ago
6

Which pair of numbers is relatively prime?

Mathematics
2 answers:
lidiya [134]3 years ago
5 0

Answer:

4 and 15

Step-by-step explanation:

i just took the quiz and i got it right :)

Musya8 [376]3 years ago
3 0

Answer:

A


Step-by-step explanation:


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5(y + 3) = 5y + ____
Sergeeva-Olga [200]

Answer:

15

Step-by-step explanation:

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3 years ago
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I'm stuck on a question I'm helping my brother with. We need to find out if y = -x squared is linear or not.
Anna71 [15]
Y=-x^2 is non linear due to the x being squared
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3 years ago
Translate the figure 5 units up. Then decide if each statement about translated figures is true or false,
m_a_m_a [10]

Answer:

First, second, and third are true.

Fourth is false.

Step-by-step explanation:

1. When we're translating a figure, everything about the figure stays the same except its location on the coordinate plane. The side lengths, the angle measures, and parallel sides will not change.

2. The fourth one is false because two figures/objects are congruent if they have the same shape and size. Since translation only affects the location on a coordinate plane, the original and final figure are congruent.

5 0
2 years ago
WILL GIVE BRAINLIEST
Serhud [2]

Answer:

5-2x=4-2x+4+x=8-x

x=5-8=-3

4 0
2 years ago
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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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