6xy (x 2 - xy + y 2 )

Answers

Answer 1

Answer:

(if 2 is a square)

6xy ( x2 - xy + y2)

= 6x3y - 6x2y2 - 6xy3 --------------the threes and twos on this line are squares and cubics. the six is just a whole number

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Answers

Answer:

Perimeter of the given figure is :---

$(2n) + (2n) +( n + 5) + (n + 5) + (3n + 3) \n = 2n + 2n + n + 5 + n + 5 + 3n + 3 \n = \boxed{9n + 13}✓$

(9n+13) is the right answer

Y = x - 2 and y=-2x +7

Answers

$\huge\boxed{\boxed{\bold{(3, 1)}}}$

$\hrulefill$

I'm assuming you need to find the solution to this system of equations (where the lines intersect).

We can use the substitution method to solve this system. Take the value of $y$ from the second equation and substitute it into the first:

$-2x+7=x-2$

Add $2x$ to both sides of the new equation:

$7=3x-2$

Now add $2$ to both sides of the equation:

$9=3x$

Divide both sides by $3$ :

$\boxed{3}=x$

Now let's solve for $y$ by substituting the known value of $x$ into the first equation:

$y=3-2$

Simplify using subtraction:

$y=\boxed{1}$

This means our solution is:

$\large\boxed{(3, 1)}$

Answer:

x = 3, y = 1

Step-by-step explanation:

Solve the following system:

{y = x - 2 | (equation 1)

y = 7 - 2 x | (equation 2)

Express the system in standard form:

{-x + y = -2 | (equation 1)

2 x + y = 7 | (equation 2)

Swap equation 1 with equation 2:

{2 x + y = 7 | (equation 1)

-x + y = -2 | (equation 2)

Add 1/2 × (equation 1) to equation 2:

{2 x + y = 7 | (equation 1)

0 x+(3 y)/2 = 3/2 | (equation 2)

Multiply equation 2 by 2/3:

{2 x + y = 7 | (equation 1)

0 x+y = 1 | (equation 2)

Subtract equation 2 from equation 1:

{2 x+0 y = 6 | (equation 1)

0 x+y = 1 | (equation 2)

Divide equation 1 by 2:

{x+0 y = 3 | (equation 1)

0 x+y = 1 | (equation 2)

Collect results:

Answer: {x = 3, y = 1

Simplify the expression.
(6+8)÷2
a.10
b.28
c.7
d.11

Answers

(6+8)÷2
(14)÷2
7

The answer is C.

(6 + 8) ÷ 2
14 ÷ 2
7
The answer is equal to C.

Write an equation for a parabola with x-intercepts (-2,0) and (5,0) which passes through the point (4,-12)