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3 years ago

In the given figure pt is the bisector of angle qpr. qs=3x, rs=2x+2, pq=3y-1 and pr=y+5

Mathematics

1 answer:

Naddika [18.5K]3 years ago

5 0

Answer:

See solutions below

Step-by-step explanation:

1) From the diagram QS = RS

3x = 2x+2

3x-2x = 2

x = 2

Since RS = 2x+2

RS = 2(2)+2

RS = 4 + 2

RS = 6

2) RQ = RS + QS

RQ = 2x+2 + 3x

RQ = 5x+2

RQ = 5(2)+2

RQ = 12

3) PQ = PR

3y-1 = y+5

3y-y = 5+1

2y = 6

y = 3

PQ = 3y-1

PQ = 3(3)-1

PQ = 8

4) PR = y+5

PR = 3+5

PR = 8

5) PQSR = PQ + QR

PQSR = 8 + 12

PQSR= 20

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Let

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For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
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$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
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So putting everything together, we found

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7 0

3 years ago

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olya-2409 [2.1K]

Answer:

angle 9 measures 105 degrees

Step-by-step explanation:

measure of angle 8 is congruent to measure of angle 11 (alternate-interior angles are congruent)

measure of angle 11 is congruent to measure of angle 9 (corresponding angles are congruent)

therefore, measure of angle 8 = measure of angle 9

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2 years ago

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telo118 [61]

When two variables are inversely proportional the relation between them can be written as:

$z=\frac{k}{x}$

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