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choli [55]
3 years ago
13

Eps in the correct order to help Kendra.

Mathematics
1 answer:
jeka57 [31]3 years ago
5 0
<h3>Answer:</h3>
  • 3,200 + 560 + 16 √ [ True ]
  • (400 × 8) + (70 × 8) + (2 × 8) √ [ True ]
  • (400 × 8) × (70 × 8) × (2 × 8) X [ False ]
  • 3,776 √ [ True ]
  • 3,200 × 560 × 16 √ [True ]
  • (400 + 70 + 2) × 8 √ [ True ]
  • (400 × 70 × 2) × 8 X [ False ]
  • 28,672,000 X [ False ]
<h3>Step-by-step explanation:</h3>

3,200 + 560 + 16 √

(400 × 8) + (70 × 8) + (2 × 8) √

(400 × 8) × (70 × 8) × (2 × 8) X

3,776 √

3,200 × 560 × 16 √

(400 + 70 + 2) × 8 √

(400 × 70 × 2) × 8 X

28,672,000 X

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In order to easily discern which graph is a proper representation of 6x + 4y = 8, you first need to convert the equation to y = mx+ b, also known as slope-intercept form. Here's how you can do this:
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4y = -6x + 8
y = -1.5x + 2
The +2 tells you that your line will intercept the vertical y-axis at (0, 2). This narrows it down to graphs a and d. Then, because you have a NEGATIVE number in front of your x (it's -1.5), you can tell that your graph will be going down as it moves from left to right. This leaves you with graph d as your answer!
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2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}

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(x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)

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H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}

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• if the Hessian determinant is negative at a given critical point, then you have a saddle point

• if both the determinant and h_{xx} are positive at the point, then it's a local minimum

• if the determinant is positive and h_{xx} is negative, then it's a local maximum

• otherwise the test fails

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\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0

while

\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0

So, we end up with

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