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3 years ago

Use the quadratic formula to solve x^2-2x+2=0

1 answer:

3 0

The quadratic formula is...
$\frac{-b (+/-)\sqrt{b^2-4ac}}{2a}$
your a will be in front of your x^2, b in front of your x, and c will be the number without an x. Plug those into the equation and you'll get two answers. That will be it

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A major car manufacturer wants to test a new engine to determine if it meets new air pollution standards. The mean emission, μ,

bonufazy [111]

Answer:

Step-by-step explanation:

the appropriate null and alternative hypotheses's are something

8 0

2 years ago

What is the perimeter of 16cm 25cm

Cloud [144]

Im guessing this is a rectangle? If so it would be
16+16+25+25
32+50
82 is the perimeter

4 0

3 years ago

Read 2 more answers

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

HELP MATH ONE MORE AFTER THIS

Lady_Fox [76]

Answer:

58 units squared

Step-by-step explanation:

We want to find the area of the square. To do so, we need to find the hypotenuse of the right triangle because this coincides with the side length of the square.

We use the Pythagorean Theorem, which states that for a right triangle with legs a and b and hypotenuse c:

a^2 + b^2 = c^2

Here, a = 7 and b = 3, so:

7^2 + 3^2 = c^2

c^2 = 49 + 9 = 58

Now, the area of a square is: A = s^2, where s is the side length. Well, c is the side length, and we've already found what c^2 is (it's 58), so that means the area of the square is 58 units squared.

Thus, the answer is 58 units squared.

5 0

3 years ago

Read 2 more answers

I'm really confused, can someone please help me with this problem?

stiv31 [10]

Answer: x = 6.25

5 ÷ 0.8

= 6.25

6 0

3 years ago