Answer:
24
Step-by-step explanation:
:)
<h3>
Answer: sometimes true</h3>
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Explanation:
The plane P can be thought of a perfectly flat ground. Now imagine a flag pole which represents line GH. If AB is drawn in chalk on the pavement, and this line AB intersects the base of the flagpole, then we've made AB and GH intersect. However, this example shows that GH is <u>not</u> on the plane P.
Is it possible to have GH be in the the plane? Yes. We could easily draw another chalk line on the ground to have it intersect AB somewhere. But as the previous paragraph says, it's also possible that GH is not in the plane.
Therefore, the statement is sometimes true
69. The jacket is said to have an original price of 45.00 dollars.
Now, is was given a 25% off.
Solve for the final cost of the jacket after the discount is applied.
=> 25% = 25% / 100% = 0.25
=> 45.00 * .25 = 11.25 dollars is the discount.
Now, let’s subtract the value of 25% by the original value.
=> 45 – 11.25 = 33.75 dollars.
This is now the costs of the jacket after the discount is applied.
Y=-6(10)+15
y=-60+15
y=-45
y=-120+15
y=-105
range equals -105 to 45
Taking
and differentiating both sides with respect to
yields
![\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](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have
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](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have
Now, when
and
, we have
