All categories

2 years ago

Quadrilateral ABCD is located at (-8,9),(-8,-5), (9,-3), and (5,4).

Mathematics

1 answer:

DENIUS [597]2 years ago

3 0

Answer:

The answer is D, i took the test and made a 100

Step-by-step explanation:

You might be interested in

WILL GIVE BRAINLIEST ANSWER !! NEED HELP QUICK !!

k0ka [10]

Answer:

its letter A

Step-by-step explanation:

8 0

2 years ago

he wholesale club sells a large box of crackers with 12 reams of crackers for $9.89. Your local store’s box of crackers has 4 re

Lostsunrise [7]

You save $2.98. You can get the same amount from your local store by buying 3 boxes, but it will cost $12.87, and if you wanna buy the same amount from the other store, you can save $2.98, since it is the same amount and costs $9.89. Hope this helps, and good luck!!

8 0

3 years ago

the height of two is recorded after 1, 2, and 3 weeks as shown in the graph at the right. which plants' growth represents a prop

vagabundo [1.1K]

For the more amount of the time the larger it will grow.

6 0

2 years ago

Question is in the picture and answer

dmitriy555 [2]

Answer:

The correct answer is D) y + 1 = -2(x - 5)

Step-by-step explanation:

All equations can be written in the following point-slope form.

y - y1 = m(x - x1)

In which, m is the slope. (x1, y1) is the given point. If we use the information given we can find the equation.

y - - 1 = -2(x - 5)

y + 1 = -2(x - 5)

8 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

2 years ago