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8_murik_8 [283]
3 years ago
15

HELP I HAVE A QUICK CHECK I HATE DIS BEACH

Mathematics
2 answers:
yawa3891 [41]3 years ago
8 0

Answer:

D. 6/8

Step-by-step explanation:

belka [17]3 years ago
4 0

Answer:

2/8

Step-by-step explanation:

1 x 1 = 2

2 x4 = 8

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Charisse is purchasing a leather handbag for $123.85, including tax. She gives the sales clerk 1 fifty-dollar bill, 2 twenty-dol
jek_recluse [69]
6.15    hope this helped 

7 0
3 years ago
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The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell
emmasim [6.3K]

Answer:

a)

Highest (-3,-3)

Lowest (2,2)

b)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12

completing the squares:

\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of  

\bf f(x,y)=x^2+y^2

subject to the constraint

\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0

Now we have

\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)

Let \bf \lambda be the Lagrange multiplier.

The maximum and minimum must occur at points where

\bf \nabla f=\lambda\nabla g

that is,

\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y

Replacing in the constraint

\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2

from this we get

<em>x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3 </em>

<em> </em>

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

b)

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0
3 years ago
What is the equation for the points (-3,1) and (9,7)
Scorpion4ik [409]

Answer:

The equation of line with given points is 2Y - X - 5 = 0

Step-by-step explanation:

Given points are ( - 3 , 1)   and   (9 , 7)

Equation of line is y = mx +c

where m is the slop of line

Now m = \frac{y2 - y1}{x2 - x1}

Or,    m = \frac{7 - 1}{9 + 3}

so, slop = \frac{6}{12}

∴   slop = \frac{1}{2}

Now the equation of line with points ( -3 , 1) and slop m is :

Y - y1 = m ( X - x1)

Or, Y - 1 =  \frac{1}{2} (X + 3)

Or, 2Y - X - 5 = 0

Hence The equation of line with given points is 2Y - X - 5 = 0 Answer

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So you would do
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5×4=20
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Patrick buys 12 bunches of bananas for $9 what will he pay for 8 bunches of bananas
larisa [96]
12b.......9\$ \\ 8b.........x \\\\ x=\frac{8*9}{12}=\frac{72}{12} \\\\ \boxed{x=5\$}
5 0
3 years ago
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