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SashulF [63]
3 years ago
9

Choose the polynomial that is written in standard form. (1 point)

Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0
The answer is D.) -2x^5y^3 + 4x^2y^2 + x^2
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Factor each expression 36abc +54ad
olya-2409 [2.1K]

Answer:

18a(2bc + 3d)

Step-by-step explanation:

36abc + 54ad

Step 1: Find the Highest Common Factor of each

36abc = 2×2×3×3×a×b×c = 18a × 2bc

54ad = 2×3×3×3×a×d = 18a × 3d

HCF = 2×3×3×a = 18a

Step 2: Factor out with HCF

18a(2bc + 3d)

7 0
3 years ago
What is the y-intercept of y = -4/3x + 4?
Reika [66]
Hello : 
<span>the y-intercept of y = -4/3x + 4 when x=0 : y= (-4/3)(0)+4=4
Answer : </span><span>A) (0, 4)</span>
7 0
2 years ago
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
X+y=1 If x=0, then y=
Ber [7]

Answer:

1

Step-by-step explanation:

x+y=1 if x=0, then y=1

5 0
2 years ago
Can someone please help asap thank you
Bumek [7]

Answer:

7%

Step-by-step explanation:

.07x1 is .07, therefore tax rate

please give brainliest im desperate

3 0
3 years ago
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