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Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The product of $\left( { 23 x }^{ 2 }{ y }^{ 2 }z \right)$ and $\left( -15{ x }^{ 3 }{ yz }^{ 2 } \right) $ is .........................  .

  1. $-345 { x }^{ 5 } { y }^{ 3 } { z }^{ 3 }$
  2. $345 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$
  3. $145 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$
  4. $170 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\left(23{x}^{2}{y}^{2}z\right)\times\left(-15{x}^{3}y{z}^{2}\right)$
$=-345‬{x}^{2+3}{y}^{2+1}{z}^{1+2}$
$=-345{x}^{5}{y}^{3}{z}^{3}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The number of integers (positive, negative or zero) solutions of
$xy-6(x+y)=0$ with x is less than or equal to y is:

  1. 5

  2. 10

  3. 12

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rewrite as xy - 6x - 6y = 0, which factors to (x - 6)(y - 6) = 36. We need pairs (x-6, y-6) such that their product is 36 and x <= y. Possible pairs (u, v) where u*v=36 and u <= v: (-36, -1), (-18, -2), (-12, -3), (-9, -4), (-6, -6), (1, 36), (2, 18), (3, 12), (4, 9), (6, 6). There are 10 such pairs.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $(8x)^2 + (6x)^2 = d^2$ and $d = 200$, then $8x \times 6x$ is equal to

  1. 18,200

  2. 18,500

  3. 18,900

  4. 19,200
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(8x)^2+(6x)^2=d^2$
or $64x^2+36x^2=d^2$
or $100x^2=d^2$
or $d=\sqrt {100x^2}=10x$
$10x=200$
$\therefore x=\frac {200}{10}=20$
Hence, $8x\times 6x=8\times 20\times 6\times 20=19,200$