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2 years ago

15

(x + y + z = 3 -y = x 2z + 3y = 0 What does x and y equal?

1 answer:

NeTakaya2 years ago

7 0

Answer:

x=2 and y=-2

Step-by-step explanation:

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To solve the system of linear equations 3 x minus 2 y = 4 and 9 x minus 6 y = 12 by using the linear combination method, Henry d

KonstantinChe [14]

It’s D my guy because of the two equations overlap each other

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2 years ago

Part A: Factor 3x2y2 − xy2 − 2y2. Show your work. (4 points)

Rudik [331]

A.

$3x^2y^2-xy^2-2y^2=y^2(3x^2-x-2)=y^2(3x^2+2x-3x-2)\\\\=y^2[x(3x+2)-1(3x+2)]=y^2(3x+2)(x-1)$

B.

$x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2=(x+3)(x+3)\\\\\text{Used}\ (a+b)^2=a^2+2ab+b^2$

C.

$x^2-25=x^2-5^2=(x-5)(x+5)\\\\\text{Used}\ a^2-b^2=(a-b)(a+b)$

6 0

2 years ago

Please don't answer wrong.

Tamiku [17]

Answer:

<h3><u>Part (a)</u></h3>

<u />

<u>Equation of a circle</u>

$(x-a)^2+(y-b)^2=r^2$

where:

(a, b) is the center
r is the radius

Given equation: $x^2+y^2=6.25$

Comparing the given equation with the general equation of a circle, the given equation is a <u>circle</u> with:

center = (0, 0)
radius = $\sqrt{6.25}=2.5$

To draw the circle, place the point of a compass on the origin. Make the width of the compass 2.5 units, then draw a circle about the origin.

<h3><u>Part (b)</u></h3>

Given equation: $x+y=1.5$

Rearrange the given equation to make y the subject: $y=-x+1.5$

Find two points on the line:

$x=-2 \implies -(-2)+1.5=3.5\implies (-2,3.5)$

$x=2 \implies -(2)+1.5\implies-0.5\implies (2,-0.5)$

Plot the found points and draw a straight line through them.

The <u>points of intersection</u> of the circle and the straight line are the solutions to the equation.

To solve this algebraically, substitute $y=-x+1.5$ into the equation of the circle to create a quadratic:

$\implies x^2+(-x+1.5)^2=6.25$

$\implies x^2+x^2-3x+2.25=6.25$

$\implies 2x^2-3x-4=0$

Now use the quadratic formula to solve for x:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

$\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-4)}}{2(2)}$

$\implies x=\dfrac{3 \pm \sqrt{41}}{4}$

To find the coordinates of the points of intersection, substitute the found values of x into $y=-x+1.5$

$\implies y=-\left(\dfrac{3 + \sqrt{41}}{4}\right)+1.5=\dfrac{3-\sqrt{41}}{4}$

$\implies y=-\left(\dfrac{3 - \sqrt{41}}{4}\right)+1.5=\dfrac{3+\sqrt{41}}{4}$

Therefore, the two points of intersection are:

$\left(\dfrac{3 + \sqrt{41}}{4},\dfrac{3-\sqrt{41}}{4}\right) \textsf{ and }\left(\dfrac{3 - \sqrt{41}}{4},\dfrac{3+\sqrt{41}}{4}\right)$

Or as decimals to 2 d.p.:

(2.35, -0.85) and (-0.85, 2.35)

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1 year ago

How do I 'simplify the square root of 50xy^2?

insens350 [35]

the square and the square root should cancel so your left with only 50xy

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