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Wittaler [7]
3 years ago
6

Please someone help me I don’t feel like doing this

Mathematics
1 answer:
Margaret [11]3 years ago
6 0

Answer:

I took a pic, if it helps

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A number, y, is equal to the difference of a larger number and 3. The same number is one third of the sum of the larger
Mnenie [13.5K]

Answer:

  • Set the larger number as x

"<em>A number, y, is equal to the difference of a larger number and 3.</em>"

y=x-3

<u>You can rearrange this equation to find the matching answer choice:</u>

y-x=-3\\x-y=3

"<em>The same number is one third of the sum of the larger number and 9.</em>"

y=\frac{1}{3} (x+9)

<u>You can rearrange this equation to find the matching answer choice:</u>

y=\frac{1}{3} x+3\\3y=x+9\\3y-x=9\\x-3y=-9

5 0
3 years ago
Help giving brainilest, heart, and 5 stars ​
PIT_PIT [208]

Answer:

4 = 6

5 = -17

Step-by-step explanation:

4. a² - b / b² - c

a² = 2² = 4

b = -2

4 - (-2) = 4 + 2 = 6

6 / b² - c = 6/4 - 3 = 6/1 = 6

4 = 6

5. -3x² + 2xy + 7

-3x² = -3 * -2² = -3 * 4 = -12

2xy = 2 * -2 * 3 = -4 * 3 = -12

-12 + -12 + 7 = -24 + 7 = -17

5 = -17

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

5 0
3 years ago
Im looking for new friends. That's me in the pic
Fantom [35]

Answer:

HI ,,,,,

I AM THERE?....

6 0
3 years ago
Let the (x; y) coordinates represent locations on the ground. The height h of
grigory [225]

The critical points of <em>h(x,y)</em> 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 <em>y</em> in the second equation and solve for <em>x</em>, then for <em>y</em> :

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. <em>h(x,y)</em> 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)}

7 0
3 years ago
HURRYYY!!! 20 POINTS AND BRAINLIEST IF CORRECT.
Gennadij [26K]

Its A i did the test

7 0
3 years ago
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