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Setler [38]
3 years ago
6

What is the quotient ? -4 /5 divide 2 A . - 1 3/5 B . -2 /5 c. 1/2 D . 1 3/ 5

Mathematics
2 answers:
DochEvi [55]3 years ago
8 0
The correct answer is b
anyanavicka [17]3 years ago
5 0

Answer:

<h3> </h3><h3>\boxed{ -  \frac{2}{5} }</h3>

Option B is the correct option.

Step-by-step explanation:

\mathrm{ -  \frac{4}{5}  \div 2}

\mathrm{dividing \: a \: negative \: and \: a \: positive \: equals \: a \: negative \:. \: ( - )  \div ( + ) = ( - )}

\mathrm{ -  \frac{4}{5}  \div 2}

\mathrm{dividing \: is \: equivalent \: to \: multiplying \: with \: the \: reciprocal}

\mathrm{ -  \frac{4}{5}  \times  \frac{1}{2} }

\mathrm{reduce \: the \: numbers \: with \: G.C.F \: 2}

\mathrm{ -  \frac{2}{5} }

Hope I helped!

Best regards!

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Leno4ka [110]

Answer:

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

Step-by-step explanation:

Given the function

y=x^2

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\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)

\mathrm{If}\:a

\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v

a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)

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Thus,

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

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