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BartSMP [9]
3 years ago
9

Find the midpoint of segment JP, given J(-4,6) and P(-12,-2).

Mathematics
1 answer:
grin007 [14]3 years ago
6 0

Answer:

The answer is

<h2>( - 8 , 2)</h2>

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

<h3>M = ( \frac{x1 + x2}{2}  ,  \:  \frac{y1 + y2}{2} )</h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

J(-4,6) and P(-12,-2)

The midpoint is

<h3>M = ( \frac{ - 4 - 12}{2}  ,  \:  \frac{6 - 2}{2} ) \\  = ( -  \frac{16}{2} ,  \:  \frac{4}{2} )</h3>

We have the final answer as

<h3>( - 8 , 2)</h3>

Hope this helps you

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Find the function y = f(t) passing through the point (0, 18) with the given first derivative.
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Answer:

\displaystyle y = \frac{t^2}{16} + 18

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<u>Algebra I</u>

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Derivative Notation

Antiderivatives - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]:                                                                   \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Property [Multiplied Constant]:                                                             \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (0, 18)

\displaystyle \frac{dy}{dt} = \frac{1}{8} t

<u>Step 2: Find General Solution</u>

<em>Use integration</em>

  1. [Derivative] Rewrite:                                                                                         \displaystyle dy = \frac{1}{8} t\ dt
  2. [Equality Property] Integrate both sides:                                                        \displaystyle \int dy = \int {\frac{1}{8} t} \, dt
  3. [Left Integral] Integrate [Integration Rule - Reverse Power Rule]:                 \displaystyle y = \int {\frac{1}{8} t} \, dt
  4. [Right Integral] Rewrite [Integration Property - Multiplied Constant]:           \displaystyle y = \frac{1}{8}\int {t} \, dt
  5. [Right Integral] Integrate [Integration Rule - Reverse Power Rule]:              \displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C
  6. Multiply:                                                                                                             \displaystyle y = \frac{t^2}{16} + C

<u>Step 3: Find Particular Solution</u>

  1. Substitute in point [Function]:                                                                         \displaystyle 18 = \frac{0^2}{16} + C
  2. Simplify:                                                                                                             \displaystyle 18 = 0 + C
  3. Add:                                                                                                                   \displaystyle 18 = C
  4. Rewrite:                                                                                                             \displaystyle C = 18
  5. Substitute in <em>C</em> [Function]:                                                                                \displaystyle y = \frac{t^2}{16} + 18

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Integration

Book: College Calculus 10e

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