Please help........................

Mathematics

2 answers:

Rama09 [41]3 years ago

8 0

Answer:

x = 7

Step-by-step explanation:

Since BD is the angle bisector of ∠ABC, then

∠ABD = ∠CBD, that is

2x + y = 14 + y ( subtract y from both sides )

2x = 14 ( divide both sides by 2 )

x = 7

N76 [4]3 years ago

3 0

Answer: x = 7, y = 2

Step-by-step explanation:

Since BD is an angle bisector, it cuts the angle into two congruent (equal) angles.

∠ABD ≅ ∠CBD

2x + y = 14 + y

2x = 14 subtracted y from both sides

x = 7 divided both sides by 2

Bonus: Let's solve for y

AD = CD

5x - y = x + 13y

5(7) - y = (7) + 13y

35 - y = 7 + 13y

28 - y = 13y subtracted 7 from both sides

28 = 14y added y to both sides

2 = y divided both sides by 14

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Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

son4ous [18]

The chain rule states that

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

Since $\ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2)$ , you have

$\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}$

and

$\dfrac{\mathrm dx}{\mathrm dt}=\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t$

Also, since $x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t$ , the derivative is

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\tan t$