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3 years ago

Which expression is equivalent to -6 - ( -2 ) PLEASE HELP MEEE

Mathematics

1 answer:

katrin [286]3 years ago

5 0

Answer:

-6 +2 = -4

Step-by-step explanation:

Its simple. Whenever you see two negative signs next to each other, they merge into an addition sign "positive".

-6+2 = -4

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If f(x)= 3x-1 and g(x)=2x-3, for which value of x does g(x)= f(2)

gregori [183]

Answer:

g(4)=f(2)

Step-by-step explanation:

f(2)=3(2)-1

f(2)=6-1

f(2)=5

We then get:

g(x)=f(2)

subsitute 5 for f(2)

g(x)=5

now solve

5=2x-3

add 3

8=2x

divide 2

4=x

so the answer is x=4 or

g(4)=f(2)

7 0

3 years ago

Sergio's internet provider charges its customers $9 per month plus 4¢ per minute of on-line usage. Sergio received a bill from t

Degger [83]

$81.40-$9.00=$72.40
$72.40/.4 (cents)= 181 minutes

8 0

3 years ago

If an angle is three times of its supplement, find the angle

zavuch27 [327]

Answer:

<h2>135° angle</h2>

Let the angle be x it's supplement = (180-x)

x = 3 (180-x)

⇒x=540−3x

⇒4x=540⇒x=135°

Step-by-step explanation:

Hope it is helpful....

5 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:

Highest (-3,-3)

Lowest (2,2)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

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

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

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

4. The sum of iwo numbers is 36. Twice the first

Ludmilka [50]

Answer:

14 and 22

Step-by-step explanation:

If x is the first number, then 36-x is the second. We are told that ...

2x -(36-x) = 6 . . . . twice the first less the second is 6

3x = 42 . . . . . . . . . add 36

x = 14

(36 -x) = 22

The numbers are 14 and 22.

4 0

3 years ago