In ΔJKL, j = 6.3 inches, l = 6.1 inches and ∠L=54°. Find all possible values of ∠J, to the nearest 10th of a degree.
1 answer:
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(a) If there is a scalar function f(x, y) such that ∇ f(x, y) = x² i + y² j, then
∂f/∂x = x²
∂f/∂y = y²
Integrating both of these equations with respect to x and y (respectively) gives
f(x, y) = 1/3 x³ + g(y)
f(x, y) = 1/3 y³ + h(x)
Differentiating with respect to the other variable gives
∂f/∂y = g'(y) = y²
∂f/∂x = h'(x) = x²
so it follows that
f(x, y) = 1/3 x³ + 1/3 y³ + C
for some constant C.
(b) By the gradient theorem,
Answer: DEb = 26°
Step-by-step explanation:
<u>Given information</u>
CEF = 7x + 21
FEB = 10x - 3
<u>Given expression deducted from the definition of the bisector</u>
FEB = CEF
<u>Substitute values into the expression</u>
10x - 3 = 7x + 21
<u>Subtract 7x on both sides</u>
10x - 3 - 7x = 7x + 21 - 7x
3x - 3 = 21
<u>Add 3 on both sides</u>
3x - 3 + 3 = 21 + 3
3x = 24
<u>Divide 3 on both sides</u>
3x / 3 = 24 / 3
x = 8
<u>Find the sum of the angle of CEF and FEB</u>
7x + 21 + 10x - 3
=7 (8) + 21 + 10 (8) - 3
=56 + 21 + 80 - 3
=77 + 80 - 3
=157 - 3
=154
<u>Subtract 154 from the straight angle</u>
DEB = 180 - 154
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