Given the area, to rewrite the function using distributive property we get:
6(2x+5y)
using distributive property we open the parenthesis, this will give us:
6×2x+6×5y
simplifying the above we get
=12x+30y

Answer: 12x+30y

6 0

3 years ago

Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

3 years ago

the sum of three numbers is 144. the third number is 4 times the first. the second number is 6 more than the first. what are the

Law Incorporation [45]

First I would change the descriptions of the numbers into expressions.

first number is n
second number is n + 6
third number is 4n (4 x n)

Then you would insert these expressions into an equation and isolate n.

n + n + 6 + 4n = 144
n + n + 4n = 144 - 6
6n = 138
n = 138/6
n = 23

Lastly, you would plug in this value into all of the expressions.

first number is 23
second number is 23 + 6 = 29
third number is 4(23) = 92

Therefore, the numbers are 23, 29, and 92.

7 0

3 years ago

Prove that the product of any two consecutive natural numbers is even.

Vitek1552 [10]

Answer with Step-by-step explanation:

Let us assume the 2 consecutive natural numbers are 'n' and 'n+1'

Thus the product of the 2 numbers is given by

$Product=n(n+1)\\\\$