3. Write the ones digit in the square of each of the following numbers

If the parent function is \root(3)(x), describe the translation of the function y=\root(3)(x+4)-3

xeze [42]

The parent function moves 4 units right and 3 units down. This is a translation by the vector.

According to the question,

Transformation $\sqrt[3]{x}$ maps onto $\sqrt[3]{x+4} -3$ . Take f(x) = $\sqrt[3]{x}$ and f(x+4)= $\sqrt[3]{x+4}$ .

f(x) = $\sqrt[3]{x}$

f(x+4)= $\sqrt[3]{x+4}$

f(x+4) - 3= $\sqrt[3]{x+4}$ - 3

Analyze the function to see the translations. f(x-4) corresponds to a horizontal translation 4 units in the positive 'x' direction. f(x)-3 corresponds to a vertical translation 3 units in the negative 'y' direction.

Hence, the parent function moves 4 units right and 3 units down. This is a translation by the vector.

Learn more about parent function here

brainly.com/question/7154028

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7 0

2 years ago

Alex, Toby and Samuel are playing a game together.

Shalnov [3]

Answer:

1/9

Step-by-step explanation:

There are three places and three people.

Times them together for the total outcomes:

1/3 x 1/3= 1/9

8 0

3 years ago

Answer this please thanks

densk [106]

Answer:

z=5, m<b=66 degrees

Step-by-step explanation:

8z+74+2z+56=180

10z+130=180

10z=50

z=5

3 0

3 years ago

Read 2 more answers

1. What is the solution for the inequality 4x < 12?

DedPeter [7]

｡☆✼★ ━━━━━━━━━━━━━━ ☾

4x < 12

Divide both sides by 4:

x < 12/4

x < 3

That is your solution ^^

Have A Nice Day ❤

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8 0

3 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

x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3

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