{{asɛmtiNe Ɔkyerɛwfo}} | Nhomakorabea | Nneɛma a ɛma wotumi yɛ ade

1

Solve the equation.

$\frac{1}{2}(6x + 2) - 6 + x = 2x + 5$

$x=$

A403

1

If AB = CD, find the length of AB.

G202

1

Paulo picked a number. He multiplied the number by 3, added 2 to the result, multiplied that result by −4, and ended up with 52. What number did Paulo pick?

AF402

1

Find all values of $x$ so that the perimeter of the square and the rectangle are equal.

G403

1

Evaluate. $(-1)^{20} + \left(\frac{1}{2}\right)^2 + (2^3) \cdot (-6 - (-7))$

N 201

4

$\left[\begin{array}{cc} \frac{2}{3}\left(a - \frac{1}{4}\right) & 2 \\ 2c - 3 & -2(4 - d) \end{array}\right] = \left[\begin{array}{cc} a + 2 & 3\left(b + \frac{8}{3}\right) \\ 3c - 4 & -12 \end{array}\right]$

Solve for a, b, c, and d.

$a=$

$b=$

$c=$

$d=$

N 607

1

Solve. $-2(x - 4) - 3(4 - x) = -1(2 - 2x)$

$x=$

A403

Algebra Mid-Chapter 4 Quiz

Algebra Mid-Chapter 4 Quiz

If AB = CD, find the length of AB.

Paulo picked a number. He multiplied the number by 3, added 2 to the result, multiplied that result by −4, and ended up with 52. What number did Paulo pick?

Find all values of $x$ so that the perimeter of the square and the rectangle are equal.

Evaluate. $(-1)^{20} + \left(\frac{1}{2}\right)^2 + (2^3) \cdot (-6 - (-7))$

If AB = CD, find the length of AB.

Paulo picked a number. He multiplied the number by 3, added 2 to the result, multiplied that result by −4, and ended up with 52. What number did Paulo pick?

Find all values of $x$ so that the perimeter of the square and the rectangle are equal.

Evaluate. $(-1)^{20} + \left(\frac{1}{2}\right)^2 + (2^3) \cdot (-6 - (-7))$