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stealth61 [152]
3 years ago
13

Differentiate \[\ln \sin^2 x\]

Mathematics
1 answer:
iragen [17]3 years ago
7 0
Y = ln {[sin(x)]^2}

use the chain rule

y' = 1 /[sin(x)]^2 * 2sin(x)*cos(x) = 2cos(x) / sin(x) = 2 cot(x)

Answer: 2 cot(x)


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−px+r=−8x−2<br> solve for X
Arlecino [84]
-8x-2=0
-8x=2
8x=-2
x=-1/4
5 0
3 years ago
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In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH
solniwko [45]

Answer:

See explanation

Step-by-step explanation:

In ΔABC, m∠B = m∠C.

BH is angle B bisector, then by definition of angle bisector

∠CBH ≅ ∠HBK

m∠CBH = m∠HBK = 1/2m∠B

CK is angle C bisector, then by definition of angle bisector

∠BCK ≅ ∠KCH

m∠BCK = m∠KCH = 1/2m∠C

Since m∠B = m∠C, then

m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH   (*)

Consider triangles CBH and BCK. In these triangles,

  • ∠CBH ≅ ∠BCK (from equality (*));
  • ∠HCB ≅ ∠KBC, because m∠B = m∠C;
  • BC ≅CB by reflexive property.

So, triangles CBH and BCK are congruent by ASA postulate.

Congruent triangles have congruent corresponding sides, hence

BH ≅ CK.

5 0
3 years ago
The legs of a 45 45 90 triangle have a length of 8 units what is the length of the hypotenous
Crank

Answer:

11.31 units

Step-by-step explanation:

a^2 + b^2 = c^2

8^2 + 8^2 = c^2

64 + 64 = c^2

128 = c^2

Opposite of exponents is square roots, and we need to get rid of that exponent in c^2, meaning we find the square root.

\sqrt{128\\} \\\\ = c

c = 11.31 units

3 0
3 years ago
Please please help me I can figure this out :-,(
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It's around 255

I hope this helps

7 0
3 years ago
3x-7z+10 how do you simplify it.
n200080 [17]
It is already simplified to the furthest extent
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3 years ago
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