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NeX [460]
3 years ago
14

Determine the parametric equations of the position of a particle with constant velocity that follows a straight line path on the

plane if it starts at the point P(7,2) and after one second it is at the point Q(2,7).
Mathematics
1 answer:
anzhelika [568]3 years ago
7 0
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Please help explanation if possible
expeople1 [14]

9514 1404 393

Answer:

  • steeper
  • up (4 units)

Step-by-step explanation:

When the equation of the line is in the form ...

  y = mx + b

the coefficient 'm' is the slope of the line, and the value 'b' is the y-intercept, the point on the y-axis where the line crosses.

We say the line is "steeper" when the magnitude of the slope is greater. Line B, with its slope of 7 will be steeper than line A, with its slope of 3.

  the new slope is steeper

__

The y-intercept of line A is -3. The y-intercept of line B is +1, so it crosses the y-axis above the point where line A crosses. One could say that line B is shifted up, when referring to the y-intercept.

3 0
3 years ago
What is 10 times 2 divided by 4 plus 3
PtichkaEL [24]
The answer would be 8!!
4 0
3 years ago
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Which number produces an irrational number when multiplied by 0.4?
snow_lady [41]

Answer:

c

Step-by-step explanation:

3 0
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Determine if the statement below makes sense. Explain.
Katarina [22]

Answer:

c

Step-by-step explanation:

$376.35 / 4 does equal to $94.0875 but when it comes to money, it should be rounded to the nearest cent being $94 and .08 cents.

4 0
3 years ago
Find the inverse of f(x)=1/(x^3)
Pepsi [2]

Answer:

Step-by-step explanation:

yf(x)^{-1}  = inverse\\f(x)=y \\y = 1/(x^{3} \\Inverse: y=x ------------> x = 1/y^{3}\\y^{3} - \frac{1}{x} = 0\\y^{3} = \frac{1}{x}\\y = \sqrt[3]{\frac{1}{x}} \\y = \frac{\sqrt[3]{1} }{\sqrt[3]{x}} \\y = \frac{1}{\sqrt[3]{x}}

4 0
3 years ago
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