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statuscvo [17]
3 years ago
5

A. 4 b. 5 c. 9 d. 25

Mathematics
1 answer:
trapecia [35]3 years ago
7 0
5t=45
45/5=9

The answer is C
Hope I helped
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Use the given equation to find the number <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20x%20-%2019.7%20%3D%20
ziro4ka [17]

Answer:

x = 31.2

Step-by-step explanation:

\frac{1}{2}x - 19.7 = -4.1

\frac{1}{2}x = 15.6 --- add 19.7 to both sides to get rid of -19.7

x = 31.2 --- multiply both sides by 2 to get rid of \frac{1}{2}

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3 years ago
Consider the equation 2x+1=16
Aliun [14]

Step-by-step explanation:

2x + 1 = 16

2x = 15 (subtract 1 on both sides)

x = 7.5 (divide 2 om both sides)

Since 7.5 is not an integer, (a) is No.

However 7.5 is rational, therefore (b) is Yes.

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3 years ago
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maw [93]

Answer:

A, B, C, D, and F are all solutions.

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6 0
3 years ago
7.2 is 250% of what number
Vlada [557]

Answer:

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2 years ago
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Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2
Mkey [24]

For this case, we have to:

By definition, we know:

The domain of f (x) = \sqrt [3] {x} is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

y = \sqrt [3] {x-2} withx = 0: y = \sqrt [3] {- 2} is defined.

y = \sqrt [3] {x+2}with x = 0:\ y = \sqrt [3] {2} is also defined.

f (x) = \sqrt {x}has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

y = \sqrt {x-2} is not defined, the term inside the root is negative whenx = 0.

While y = \sqrt {x+2} if it is defined for x = 0.

Answer:

y = \sqrt {x-2}

Option b

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