$\dfrac{5+2\sqrt3}{7+4\sqrt3}\qquad\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{5+2\sqrt3}{7+4\sqrt3}\cdot\dfrac{7-4\sqrt3}{7-4\sqrt3}=\dfrac{(5+2\sqrt3)(7-4\sqrt3)}{7^2-(4\sqrt3)^2}\qquad\text{use FOIL}\\\\=\dfrac{(5)(7)+(5)(-4\sqrt3)+(2\sqrt3)(7)+(2\sqrt3)(-4\sqrt3)}{49-4^2(\sqrt3)^2}\\\\=\dfrac{35-20\sqrt3+14\sqrt3-8(3)}{49-(16)(3)}=\dfrac{(35-24)+(-20\sqrt3+14\sqrt3)}{49-48}\\\\=\dfrac{11-6\sqrt3}{1}=11-6\sqrt3$

8 0

3 years ago

Tell whether the ordered pair is a solution of the system of equations.

katovenus [111]

Answer:

2x + y = 5

y = 4x – 1

Substitute (2, 1) into system, then:

2x + y = 5 => 2 x 2 + 1 = 5 => 5 = 5 (satisfied)

y = 4x– 1 => 4 x 2 - 1 = 1 => -7 = 1 (not satisfied)

=> Option 3 is correct.

Hope this helps!

8 0

3 years ago

The rectangle has an area of 4(x+3) square units.

Serggg [28]

Suppose there is a rectangle with a width and length of x and y. The area would be xy.

A rectangle with width and length 2x and 2y would have an area of 4xy.

In essence, when you multiply the lengths by 2, the area multiplies by 2 SQUARED.

4(4(x+3)) = 16(x+3).

The ratio will be the same for any value of x because the area is still being multiplied by the same 2².

5 0

3 years ago