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3 years ago

A- (2, -3) B- (2, -1) C- (5, -3) D- (7, -1) PLEASE HELP!!!

Mathematics

1 answer:

Aneli [31]3 years ago

3 0

Answer:

Step-by-step explanation:

Line segment divided into 5:3 ratio, so one segment is ⅝ of AB and one segment is ⅜ of AB.

Compare the coordinates of A and B.

change in x-coordinates = 10-2 = 8

change in y-coordinates = 2-(-6) = 8

⅝ of 8 = 5, so x-coordinate of C = (x-coordinate of A) + 5 = 7, and y-coordinate of C = (y-coordinate of A) + 5 = 1

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I WILL MARK BRAINLIEST!!

Stels [109]

Answer:

9 inches

Step-by-step explanation:

The population is within 6 standard deviations of the mean with a confidence of 99.7 percent, 3 above and 3 below

The standard deviation of the sample mean is equal to the population standard deviation divided by the square root of the sample size

population standard deviation = 36

sample size = 144

standard deviation of the sample mean = 36/ sqrt(144)

=36/12

The margin of error is the number of standard deviations above the mean * the standard deviations

margin of error = 3 * 3

= 9 inches

8 0

3 years ago

Let the (x; y) coordinates represent locations on the ground. The height h of

grigory [225]

The critical points of h(x,y) occur wherever its partial derivatives $h_x$ and $h_y$ vanish simultaneously. We have

$h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5$

Substitute y in the second equation and solve for x, then for y :

$2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}$

This is to say there are two critical points,

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

To classify these critical points, we carry out the second partial derivative test. h(x,y) has Hessian

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

whose determinant is $192y-16$ . Now,

• 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

We have

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

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

$h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}$