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brilliants [131]
3 years ago
8

Point P is the centroid of LMN. Find PN

Mathematics
1 answer:
kolbaska11 [484]3 years ago
8 0

Answer:

PN = 14

QP = 7

Step-by-step explanation:

Extend PQ to R so that QR=PQ

MQ = QL  (known, mid-point)

MPLR is parallelogram (diagonal bisect each other)

MP // RL          NP/PR = NE/EL = 1/1 (E mid-point, known)

NP = PR

PQ = QR = 1/2 PR = 1/2 NP

NP = 2 PQ

NP + PQ = 2PQ + PQ = 3 PQ = QN = 21

PQ = 7

PN = 14

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Which of the following shows that polynomials are closed under addition when two polynomials 4x2 − 8x − 7 and −5x + 16 are added
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First we have to combine like terms. 4x² is one term. -8x is another term. -7 is another term. poly = "many" ⇒ polynomial = many terms

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Combine like terms.
Step 1: 4x² has no other terms with x² so it stays by itself.
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

y = 7x - 4x² 

<span>7x - 4x² = 0 </span>

<span>x(7 - 4x) = 0 </span>

<span>x = 0, 7/4 </span>

<span>Find the area of the bounded region... </span>

<span>A = ∫ 7x - 4x² dx |(0 to 7/4) </span>

<span>A = 7/2 x² - 4/3 x³ |(0 to 7/4) </span>

<span>A = 7/2(7/4)² - 4/3(7/4)³ - 0 = 3.573 </span>

<span>Half of this area is 1.786, now set up an integral that is equal to this area but bounded by the parabola and the line going through the origin... </span>

<span>y = mx + c </span>

<span>c = 0 since it goes through the origin </span>

<span>The point where the line intersects the parabola we shall call (a, b) </span>

<span>y = mx ===> b = m(a) </span>

<span>Slope = m = b/a </span>

<span>Now we need to integrate from 0 to a to find the area bounded by the parabola and the line... </span>

<span>1.786 = ∫ 7x - 4x² - (b/a)x dx |(0 to a) </span>

<span>1.786 = (7/2)x² - (4/3)x³ - (b/2a)x² |(0 to a) </span>

<span>1.786 = (7/2)a² - (4/3)a³ - (b/2a)a² - 0 </span>

<span>1.786 = (7/2)a² - (4/3)a³ - b(a/2) </span>

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<span>1.786 = (7/2)a² - (4/3)a³ - (7/2)a² + 2a³ </span>

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<span>Slope = m = b/a = 2.00 / 1.39 = 1.44</span>

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