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Liula [17]
3 years ago
14

1. ∠ PRS and ∠ RPQ are right angles

Mathematics
2 answers:
Blababa [14]3 years ago
7 0

Answer:

Given : PS = QR

            ∠PRS = ∠RPQ

            PR = PR (On the same base)

Proof / Reason :

∠PRS = ∠RPQ (R)

PS = QR          (H)

PR = PR           (S)

Therefore, Triangle PRS is congruent to Triangle RPQ

∠PSR = ∠RQP     By  C.P.C.T. ( Congruent Parts Of Congruent Triangles)

Thanks!


natima [27]3 years ago
7 0

Answer:

the other answer is wrong. that is not the correct answer. dont give it a 5 star and dont like it.

Step-by-step explanation:

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Answer:

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

General Formulas and Concepts:

<u>Pre-Algebra</u>

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<u>Algebra I</u>

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Derivative Notation

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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

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Unit: Integration

Book: College Calculus 10e

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