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sergiy2304 [10]
3 years ago
5

In which form is the following function written? y+20=8(x+7)^2

Mathematics
1 answer:
Wewaii [24]3 years ago
7 0

Answer:

  (a variation of) vertex form

Step-by-step explanation:

For vertical scale factor "a" and vertex (h, k), the vertex form of the equation for a parabola can be written as ...

  y = a(x -h)^2 +k

If k is subtracted from this equation, an alternate form is ...

  y -k = a(x -h)^2

This latter version of vertex form is the form your equation has, where ...

  • a = 8
  • (h, k) = (-7, -20)

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Step-by-step explanation:

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3 years ago
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Llana [10]

Answer:

\displaystyle log_\frac{1}{2}(64)=-6

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

log_b(1) = 0

Logarithm of the base:

log_b(b) = 1

Product rule:

log_b(xy) = log_b(x) + log_b(y)

Division rule:

\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)

Power rule:

log_b(x^n) = n\cdot log_b(x)

Change of base:

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Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

\displaystyle log_\frac{1}{2}(64)

Factoring 64=2^6.

\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)

Since

\displaystyle 2=(1/2)^{-1}

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})

Applying the logarithm of the base:

\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}

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Step-by-step explanation:

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16/25 is 0.64 as a fraction in lowest terms

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