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madam [21]
3 years ago
10

Two factors are multiplied and their product is 34.44 one factor is a whole number what is the least number of decimal places in

the other factors explain
Mathematics
1 answer:
katovenus [111]3 years ago
7 0

Answer:

2 decimal places

Step-by-step explanation:

When you multiply two decimals, the largest possible number of decimal places in the product is the sum of the number of the decimal places of the factors. Here, the product has 2 decimal places. One factor is an integer, so it has no decimal places. The other factor must have at least 2 decimal places.

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To solve the inequality \frac{x^2+x+3}{2x^2+x-6}\ge \:0

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so factor the denominator 2x^2+x-6:

2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\

Now take \mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)

Then we get factor of the denominator as \left(2x-3\right)\left(x+2\right)

Thus \frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}

Now \mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}\\

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\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0

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