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Katyanochek1 [597]
3 years ago
9

J is the midpoint of hk. what is hj, jk, and hk

Mathematics
2 answers:
Lyrx [107]3 years ago
4 0
Hj and jk are the same length line segments ( because the midpoint divides a line into two equal parts)
So hj = jk.

hk is the line segment which has the mid point j. It is the double of hj or jk. It can be the sum of hj and jk. 
hj + jk = hk 
      or
2 * hj = hk
      or 
2 * jk = hk
kari74 [83]3 years ago
4 0

Answer:

The answer is 'b'  on Edg

Step-by-step explanation:

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\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

Now, when x=1 and y=2, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
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