She saves 1/10 of her salary. That's the simplified version, it can also be written 10/100. Fractions, decimals, and percentages are all interchangeable.

5 0

3 years ago

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If ΔRED ≅ ΔTAN, then which of the following statements is not true? RE ≅ TA ∠E ≅ ∠A RD≅ TA ∠D ≅ ∠N

Assoli18 [71]

Answer:

RD≅ TA

Step-by-step explanation:

RD≅ TA is not true statement as they are not corresponding sides.

4 0

3 years ago

Given that f(x) = 6x + 2 and g(x) = solve for g(f(1)).

bixtya [17]

$f(x)=6x+2\\\\
g(f(1))=6*1+2=6+2=8\\\\
Solution\ is \ g(x)=8$

4 0

3 years ago

(x^2 - x^(1/2))/(1-x^(1/2))

Levart [38]

$\frac { \left( { x }^{ 2 }-{ x }^{ \frac { 1 }{ 2 } } \right) }{ \left( 1-{ x }^{ \frac { 1 }{ 2 } } \right) }$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot 1$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot \frac { \left( 1+\sqrt { x } \right) }{ \left( 1+\sqrt { x } \right) }$

$\\ \\ =\frac { { x }^{ 2 }+{ x }^{ 2 }\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }$

$\\ \\ =\frac { -\sqrt { x } \left( 1-{ x }^{ 2 } \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\\ \\ =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\\ \\ =\frac { \left( 1-x \right) \left\{ -\sqrt { x } \left( 1+x \right) -x \right\} }{ \left( 1-x \right) }$

$\\ \\ =-\sqrt { x } \left( 1+x \right) -x\\ \\ =-{ x }^{ \frac { 1 }{ 2 } }\left( 1+{ x }^{ \frac { 2 }{ 2 } } \right) -x$

$\\ \\ =-{ x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } }-x\\ \\ =-\sqrt { x } -\sqrt { { x }^{ 3 } } -x$

3 0

3 years ago

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