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Aloiza [94]
2 years ago
6

Please answer the question correctly

Mathematics
1 answer:
WINSTONCH [101]2 years ago
6 0

Answer: The answer is 16

Step-by-step explanation: You divide the whole line segment by two to get only the half of the line segement.

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The GCF of an odd number and an even number is always even
Oxana [17]
The answer is No it is not
7 0
3 years ago
(PLEASE HELP ASAP)
lesya [120]

Answer:

  • Slope: -2
  • equation: y = -2x +5

Step-by-step explanation:

The given point is the y-intercept of the line, so the slope-intercept form of the equation of a line can be used:

  y = mx + b . . . . for slope m and y-intercept b

Here, you have the slope given as ...

  m = -2

  y-intercept = 5

so the equation of the line is ...

  y = -2x +5

_____

The y-intercept is the value of y when x=0. The point (x, y) = (0, 5) tells you the y-intercept is 5.

7 0
3 years ago
How do I solve this?
aleksandr82 [10.1K]
1) -2 -6
2) -2 -y
3) -2 -6 + -y
4) -2 -6 -y
5 0
3 years ago
Read 2 more answers
Don Price bought a house for $70,000. He rents the house for $900 per month. He has $700 in monthly expenses.
frutty [35]
Annual rent=$10,800
Annual Expenses=$8,400
Rate of income=moneyt
3 0
3 years ago
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For how many real values of x is <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B120-%5Csqrt%7Bx%7D%7D%20" id="TexFormula1" title
leva [86]

A good place to start is to set \sqrt{x} to y. That would mean we are looking for \sqrt{120-y} to be an integer. Clearly, y\leq 120, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since \sqrt{x} is a radical, it only outputs values from [0,\infty], which means y is on the closed interval: [0,120].

With that, we don't really have to consider y anymore, since we know the interval that \sqrt{x} is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval [0,120], which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

\sqrt{120-\sqrt{x}}=k \implies \\ \sqrt{x}=k^2-120 \implies\\ x=(k^2-120)^2

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.

5 0
3 years ago
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