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sukhopar [10]
3 years ago
7

The sum of fourteen and a number added to the product of thirteen and the number is

Mathematics
1 answer:
nika2105 [10]3 years ago
4 0
The answer would be -1

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1) What is the height of the parallelogram? *<br> 7 cm/<br> 5 cm<br> 15 cm
Natali5045456 [20]

Answer:

15

Step-by-step explanation:

5 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
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astraxan [27]

Answer:

16 × <u><em>X </em></u>+ 3=

Step-by-step explanation:

u will get your answer

4 0
3 years ago
A farmer has a 100 ft by 200 ft rectangular field that he wants to increase by 15.5% by cultivating a strip of uniform width aro
Advocard [28]

Answer:

(a) The strip should be 5ft wide

(b) The strip around the outside field is 10ft wide.

Step-by-step explanation:

Given:

Length of the rectangular field, L= 200 ft

width of the rectangular field, w = 100 ft

Area of the rectangular field, A = 200ft x 100ft = 20000 ft^2

let the width of the strip = x

The strip around the outside field = 2x

If the field is increased by 15.5%

New area of the field = 1.155 x 20000 = 23,100 ft^2

The increase in area of the field = 3,100 ft

3,100 = New area of field - old area of the field

3100 = (200 + 2x)(100 + 2x) - 20000

3100 = 20000 + 400x 200x + 4x^2 - 20000

3100 = 600x + 4x^2

Divide through by 4

775 = 150x + x^2

x^2 + 150x - 775 = 0

Factorize

(x + 155)(x-5) = 0

x = 5 ft

The strip should be 5ft wide.

The strip around the outside field = 2 x 5 ft = 10 ft

Thus, the strip around the outside field is 10ft wide.

3 0
3 years ago
Can y’all help me with this and you’ll get 20 points
VARVARA [1.3K]
The answers to your page the whole book is online with answer key
4 0
3 years ago
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