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Olin [163]
3 years ago
5

Someone please answer this pick the right answer

Mathematics
1 answer:
Viefleur [7K]3 years ago
5 0
770.

.........................
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Answer:

$258 <u><em>I think</em></u>

Step-by-step Explanation:

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Explain how to find the area of the triangle when x = 3 inches. then find the area
geniusboy [140]

To find the area of a triangle, multiply the base by the height, and then divide by 2. The division by 2 comes from the fact that a parallelogram can be divided into 2 triangles.

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Olenka [21]

Answer:

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Step-by-step explanation:

This equation is in "vertex form," meaning that you can identify the vertex and other features of the graph from the equation.

  y = a(x -h)² +k . . . . . the vertex is (h, k); the vertical scale factor is "a"

Comparing to your equation, you see ...

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The vertex is (h, k) = (3, -1). The vertical scale factor is negative.

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This tells you the graph opens downward (the scale factor is negative), and the vertex (maximum point) is below the x-axis. (It has a negative y-coordinate.)

Because it start below the x-axis and goes down from there, the graph does not intersect the x-axis. There are zero (0) x-intercepts.

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Expand log_1/2(3x^2/2) using the properties and rules for logarithms.<br><br><br> Please help.
Naily [24]

Answer:

\frac{0.1761+2\log(x)}{-0.3010}

Step-by-step explanation:

Data provided:

\log_{1/2}(\frac{3x^2}{2})

now,

we know the properties of log functions as:

1) log(AB) = log(A) + log(B)

2) \log(\frac{A}{B}) = log(A) - log(B)

3) log(xⁿ) = n × log(x)

thus,

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3x^2) - \log_{1/2}(2)

or

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3)+\log_{\frac{1}{2}}(x^2) - \log_{1/2}(2)

or

using property 3

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3)+2\log_{\frac{1}{2}}(x) - \log_{1/2}(2)

also,

\log_a(x)=\frac{\log(x)}{\log(a)}

thus,

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)}{\log(\frac{1}{2})}+2\times[\frac{\log(x)}{\log(\frac{1}{2}}]-\frac{\log(2)}{\log(\frac{1}{2})}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)+2\log(x)-\log(2)}{\log(\frac{1}{2})}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)+2\log(x)-\log(2)}{\log(1)-log(2)}

now,

log(1) = 0

log(2) = 0.3010

log(3) = 0.4771

thus,

\log_{1/2}(\frac{3x^2}{2}) = \frac{0.4771+2\log(x)-0.3010}{0-0.3010}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{0.1761+2\log(x)}{-0.3010}

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3 years ago
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