Whats 100*100??? please help

Mathematics

1 answer:

Svetllana [295]2 years ago

7 0

Answer: 100*100=10,000

thanks for the points :))

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Can someone please help me with this question?? It’s really hard and once you understand it, can you please give me an explanati

Thepotemich [5.8K]

We want to subtract 8x + 3 from -2x+5. We can create an expression to represent this.

-2x + 5 - (8x + 3).

After this, lets distribute the - sign (think of this like expanding something with -1).

-2x + 5 - 8x - 3

Lastly, we just need to combine like terms.

-2x + 5 - 8x - 3

Combine the 5 and -3 to get 2.

-2x + 2 - 8x

Combine the -2x and -8x to get -10x.

-10x + 2

The final answer to the question is therefore A.

3 0

3 years ago

Anyone can help me solve this equation using cross multiplying

Natali5045456 [20]

9514 1404 393

Answer:

x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)$

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

$3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}$

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Comment on "cross multiply"

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes what the result looks like. It doesn't adequately describe how you get there. The one and only rule in solving Algebra problems is "whatever is done to one side of the equation must also be done to the other side of the equation." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "multiply by the product of the denominators and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}$