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3 years ago

Please can you help me! z-² as a fraction

1 answer:

5 0

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
z^{-2}\implies \cfrac{1}{z^2}$

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Alex Ar [27]

64 gallons per minute

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3 years ago

8th grade teachers and students from M.I.T went on a field trip to an zoo. Tickets

alukav5142 [94]

As there are multiple answers going with trial and error seems good in this case.

5.00 + 5.00 + 5.00 + 5.00 = 20.00

3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 = 31.5

20.00 + 31.5 = 51.5

5.00 + 5.00 + 5.00 + 5.00 = 20.00

3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 = 35.00

20.00 + 35.00 = 55.00

5.00 + 5.00 + 5.00 = 15.00

3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 = 31.5

15.00 + 31.5 =46.5

Knowing that for the first one all A,B, and C were wrong we can determine that D is the answer. But wait if I made one mistake and didn't understand if $156.5 is equivalent to $156.50 that would mean A is correct.

Always recheck your answers and good luck,

Jon

6 0

3 years ago

-3 1/2 + -2 + 4 1/4 = ????

mylen [45]

Answer:

-1 1/4

Step-by-step explanation:

6 0

3 years ago

If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f

nikklg [1K]

The quadratic function given by:

$f(x)=a(x-h)^2+k, \ \ \ a\neq 0$

is in vertex form. The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h, k)$ . So:

To translate the graph of a function to the right, left, upward or downward we have:

$For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)$

By knowing this things, we can solve our problem as follows:

FIRST.

Translating 11 units to the left:

$g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2$

Then translating 5 units down:

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5$

Since the new function is fatter, the factor we need to multiply the term $(x+11)^2$ must be less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.