2f'(x)+2x(f(x))^3+3x^2(f'(x))=0
f(3)=-2, x=3
2f'(3)+2*3*-2^3+3*-2^2*f'(3)=0
2f'(3)-48+12f'(3)=0
14f'(3)=48
f'(3)=48/14=24/7
Answer:
p = (2k + 1)i + j
q = 25i + (k + 4)j
p . q = 11
Find the dot product of p and q
(2k + 1)(25) + (k + 4) = 11
50k + 25 + k + 4 = 11
51k = -18
k = - 18/51 = –6/17.✅
b) p.q = |p||q|cos∅
p= (2k + 1)i + j = [2(-6/17) + 1)i + j ]= 5/17i + j
|p| = √ (5/17)² + 1²
|p| = √314/ 17 ( The square root doesn't cover the 17) = 1.0424
q = 25i + (k + 4)j = 25i + ( -6/17 + 4)j
q = 25i + 62/17 j
|q| = √ 25² + (62/17)²
= 25.2646
Now applying them to the formula above
We already have p.q = 11 from the question.
11 = (1.0424)(25.2646)Cos∅
11 = 26.3358Cos∅
Cos∅ = 11/26.3358
Cos∅ = 0.4177
∅ = 65.31°.
This should be it.
Hope it helps
the answer to this question is C
Answer:
be brief can't see any picture
<span>GIVEN:
∠D congruent to ∠T
∠E congruent to ∠U
DE congruent to UT
answer
</span><span>ASA Postulate</span>
