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3 years ago

Why is the distributive property appropriate for two-digit multiplication?

1 answer:

4 0

Here's an example:

$25\times93=(20+5)\times(90+3)$
$25\times93=20\times(90+3)+5\times(90+3)$
$25\times93=20\times90+20\times3+5\times90+5\times3$
$25\times93=1800+60+450+15$
$25\times93=2325$

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Round 1.983 to the nearest hundredth

svet-max [94.6K]

Answer:

1.98

Step-by-step explanation:

Find the number in the hundredth place 8 and look one place to the right for the rounding digit 3. Round up if this number is greater than or equal to 5 and round down if it is less than 5.

4 0

3 years ago

A system of linear equations with fewer equations than unknowns is sometimes called an underdetermined system. Can such a system

elena55 [62]

Answer:

No, it cannot have a unique solution. Because there are more variables than equations, there must be at least one free variable. If the linear system is consistent and there is at least one free variable, the solution set contains infinitely many solutions. If the linear system is inconsistent, there is no solution.

Step-by-step explanation:

the questionnaire options are incomplete, however the given option is correct

We mark this option as correct because in a linear system of equations there can be more than one solution, since the components of the equations, that is, the variables are multiple, leaving free variables which generates more alternative solutions, however when there is no consistency there will be no solution

3 0

3 years ago

1) translation: 1 unit left

Otrada [13]

Answer:

x=(1,0)

g=(3,0)

q=(1,-2)

u=(2,-4)

Step-by-step explanation:

7 0

2 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$