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2 years ago

Which statement is true about the function f(x)= –StartRoot negative x EndRoot?

Mathematics

1 answer:

Fofino [41]2 years ago

7 0

Answer:

Step-by-step explanation:

The domain of the function is all positive real numbers and the range of the function is all negative numbers

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Someone helppppp!!!!!!

ioda

lol omg I don’t know

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2 years ago

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HELP THIS IS HARD SOMEONE pls

IrinaK [193]

Answer:

3rd option

Step-by-step explanation:

Since the centre of dilatation is at the origin then multiply the coordinates by $\frac{1}{3}$

K (- 3, 9 ) → K' (- 3( $\frac{1}{3}$ ), 9( $\frac{1}{3}$ ) ) → K' (- 1, 3 )

L (- 9, 0 ) → L' (- 9( $\frac{1}{3}$ ), 0 ( $\frac{1}{3}$ ) ) → L' (- 3, 0 )

M (2, - 8 ) → M' (2 ( $\frac{1}{3}$ ), - 8 ( $\frac{1}{3}$ ) ) → M' ( $\frac{2}{3}$ , - $\frac{8}{3}$ )

N (6, 4 ) → N' (6 ( $\frac{1}{3}$ ) , 4 ( $\frac{1}{3}$ ) ) → N' ( 2, $\frac{4}{3}$ )

7 0

2 years ago

Express your answer in scientific notation. 4.9 x 10^4 - 8200

Fiesta28 [93]

Answer:

4.0800*10^4

Step-by-step explanation:

4.9*10^4=49000

49000-8200=40800

since we can only have base numbers 1-9 in scientific notation, we need to move the decimal left 4 places:

4.0800*10^4

Hope this helps!

4 0

2 years ago

Create a one-to-one function.

Nina [5.8K]

Answer:

f(x)=x

Step-by-step explanation:

that's keeping it simple... ;-)

5 0

2 years ago

Which expression is equivalent to StartFraction b Superscript negative 2 Baseline Over a b Superscript negative 3 Baseline EndFr

nignag [31]

Answer:

$\frac{b^{-2}}{ab^{-3}}=\frac{b}{a}$

Step-by-step explanation:

Writing the description in algebraic translation

$\frac{b^{-2}}{ab^{-3}}$

so we have to find the expression which will be equal to $\frac{b^{-2}}{ab^{-3}}$ .

Considering the expression

$\frac{b^{-2}}{ab^{-3}}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}$

$\frac{b^{-2}}{b^{-3}}=b^{-2-\left(-3\right)}$

so the expression becomes

$=\frac{b^{-2-\left(-3\right)}}{a}$ ∵ $\frac{b^{-2}}{b^{-3}}=b^{-2-\left(-3\right)}$

$\mathrm{Subtract\:the\:numbers:}\:-2-\left(-3\right)=1$

$=\frac{b}{a}$

Therefore,

$\frac{b^{-2}}{ab^{-3}}=\frac{b}{a}$

7 0

3 years ago

Read 2 more answers