Can someone do these easy questions. I forgot fractions

Mathematics

1 answer:

Marat540 [252]2 years ago

6 0

1. 3/4
2. 3/4
3. 3 whole parts one missing part
4. 0.75 75% 6/8
5. It’s closer to 1 on a number line because it’s more then 1/2
6. A fraction greater could be 7/8 Bc it’s more then 6/8 and 6/8 is equal to 3/4
A fraction less then could be 1/4 Bc 1 is less then 3

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Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

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Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0