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Pie
2 years ago
6

Pls answer my question!! I'm sooo confused!!!!!!!!

Mathematics
2 answers:
solmaris [256]2 years ago
8 0

Answer:

It letter C!!

Step-by-step explanation:

oTheres no way you can write a negative twice with a top and bottom You can put the negative in the middle or one on ONE number  top or bottom but never can be both

kompoz [17]2 years ago
6 0

Answer:

C.

For A and B where the negative is doesn't matter since the decimal form of both of these fractions is equal to -0.66667 (they are equal no matter where u put the negative sign). the decimal form of - 2/3 is also -0.666667. A good rule to know is that a negative divided by a negative or a negative multiplied by a negative is always a positive. in c, -2/-3 is actually  0.66667 meaning its incorrect.

Hard to explain through writing but hoped this helped

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Rationalize the denominator of sqrt -9 / (4-7i) - (6-6i) ...?
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kondaur [170]

Answer:

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Step-by-step explanation:

In the question we're provided with an equation that is :

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And we are asked to find the solution for the equation .

<u>Solution</u><u> </u><u>:</u><u> </u><u>-</u>

<u>\longrightarrow \:  \frac{v}{7}  = 3</u>

Multiplying by 7 on both sides :

\longrightarrow \: \frac{v}{ \cancel{7} }\times  \cancel{7}  = 3 \times 7

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\longrightarrow \:  \blue{\boxed{v = 21}}

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We are verifying our answer by substituting value of v in the equation given in question :

\longrightarrow \: \frac{v}{7}  = 3

Putting value of v :

\longrightarrow \:  \cancel{\frac{21}{7}}  = 3

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\longrightarrow \:3 = 3

\longrightarrow \: L.H.S=R.H.S

\longrightarrow \: Hence , \: Verified.

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<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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Alchen [17]

Solution, \frac{-4r}{4}+\frac{5r}{3}=\frac{2r}{3}

Steps:

\frac{-4r}{4}+\frac{5r}{3}

\frac{-4r}{4},\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b},\\-\frac{4r}{4},\\\mathrm{Divide\:the\:numbers:}\:\frac{4}{4}=1,\\-r,\\-r+\frac{5r}{3}

\mathrm{Convert\:element\:to\:fraction}:\quad \:r=\frac{r3}{3},\\\frac{5r}{3}-\frac{r\cdot \:3}{3}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c},\\\frac{5r-r\cdot \:3}{3}

5r-r\cdot \:3,\\\mathrm{Add\:similar\:elements:}\:5r-3r=2r,\\\frac{2r}{3}

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